Chemical Kinetics-NCERT Solutions

 CBSE Class 12 Chemistry

NCERT Solutions
Chapter – 04
Chemicals Kinetics

---

In-text question
1. For the reaction R→P$\mathrm{R}\to \mathrm{P}$ , the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Ans. Average rate of reaction −Δ[R]Δt$-\frac{\mathrm{\Delta }[\mathrm{R}]}{\mathrm{\Delta }\mathrm{t}}$
=−[R]2−[R]1t,−t1$=-\frac{[\mathrm{R}{]}_2-[\mathrm{R}{]}_1}{\mathrm{t},-{\mathrm{t}}_1}$
=−0.02−0.0325Mmin−1$=-\frac{0.02-0.03}{25}{\mathrm{M}\mathrm{m}\mathrm{i}\mathrm{n}}^{-1}$
=−−0.0125Mmin−1$=-\frac{-0.01}{25}{\mathrm{M}\mathrm{m}\mathrm{i}\mathrm{n}}^{-1}$
=4×10−4Mmin−1$=4\times {10}^{-4}{\mathrm{M}\mathrm{m}\mathrm{i}\mathrm{n}}^{-1}$
=4×10−460Ms−1$=\frac{4\times {10}^{-4}}{60}{\mathrm{M}\mathrm{s}}^{-1}$
=6.67×10−5Ms−1$=6.67\times {10}^{-5}\mathrm{M}{\mathrm{s}}^{-1}$

---

2. In a reaction, 2A→$\to$Products, the concentration of A decreases from 0.5 mol L-1 to 0.4 mol L-1 in 10 minutes. Calculate the rate during this interval?
Ans. Average rate=−12Δ[A]Δt$=-\frac{1}{2}\frac{\mathrm{\Delta }[\mathrm{A}]}{\mathrm{\Delta }\mathrm{t}}$
=−12[A]2−[A]t2−t1$=-\frac{1}{2}\frac{[\mathrm{A}{]}_2-[\mathrm{A}]}{{\mathrm{t}}_2-{\mathrm{t}}_1}$
=−120.4−0.510$=-\frac{1}{2}\frac{0.4-0.5}{10}$
=−12−0.110$=-\frac{1}{2}\frac{-0.1}{10}$
= 0.005 mol L - 1 min - 1
= 5 ×$\times$10 - 3 Mol L- 1 min -1

---

3. For a reaction,A + B→$\to$Product; the rate law is given by, r = K[A]1/2[B]2. What is the order of the reaction?
Ans. The order of the reaction =12+2=212$=\frac{1}{2}+2=2\frac{1}{2}$
= 2.5

---

4. The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

Ans. The order of reaction is defined as the sum of the powers of concentrations In the rate law.
The rate of second order reaction can be expressed as rate = k[A]2
The reaction X → Y follows second order kinetics.
Therefore, the rate equation for this reaction will be:
Rate = k[X]2 ..............(1)Let [X] = a mol L-1 , then equation (1) can be written as:
Rate = k [a]2 = ka2
If the concentration of X is increased to three times, then [X] = 3a mol L-1
Now, the rate equation will be:
Rate = k [3a]2
= 9(ka2)
Hence, the rate of formation will increase by 9 times.

---

5. A first order reaction has a rate constant 1.15 ×$\times$ 10-3. How long will 5 g of this reactant take to reduce to 3 g?
Ans. From the question, we can write down the following information:
Initial amount = 5 g
Final concentration = 3 g
Rate constant = 1.15 ×$\times$ 10-3
We know that for a 1st order reaction,
t=2.303klog[R]0[R]$\mathrm{t}=\frac{2.303}{\mathrm{k}}\log \frac{[\mathrm{R}{]}_0}{[\mathrm{R}]}$
=2.3031.15×10−3log53$=\frac{2.303}{1.15\times {10}^{-3}}\log \frac{5}{3}$
=2.3031.15×10−3×0.2219$=\frac{2.303}{1.15\times {10}^{-3}}\times 0.2219$
= 444.38 s
= 444 s (approx.)

---

6. Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Ans. We know that for a 1st order reaction,
t1/2=0.693k${\mathrm{t}}_{1/2}=\frac{0.693}{\mathrm{k}}$
It is given that t1/2= 60 min
k = 0.693 / t1/2
= 0.693 / 60
= 0.01155 min-1
= 1.155 ×$\times$ 10 -2 min-1

---

7. What will be the effect of temperature on rate constant?
Ans. The rate constant of a reaction is nearly doubled with a 10 K rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation,
k=Ae−Ea/RT$\mathrm{k}={\mathrm{A}\mathrm{e}}^{-\mathrm{E}\mathrm{a}/RT}$
Where,
A is the Arrhenius factor or the frequency factor
T is the temperature
R is the gas constant
Ea is the activation energy.

---

8. The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.
Ans. It is given that T1 = 298 K
Therefore, T2 = (298 + 10)K= 308 K
We also know that the rate of the reaction doubles when temperature is increased by 10∘$\circ$.
Therefore, let us take the value of k1 = k and that of k2 = 2k
Also, R = 8.314 JK-1mol-1
Now, substituting these values in the equation:
logk2k1=E2.303R[T2−T1T1T2]$\log \frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=\frac{\mathrm{E}}{2.303\mathrm{R}}\left[\frac{{\mathrm{T}}_2-{\mathrm{T}}_1}{{\mathrm{T}}_1{\mathrm{T}}_2}\right]$
We get:
log2kk=E22.303×8.314[10298×308]$\log \frac{2\mathrm{k}}{\mathrm{k}}=\frac{{\mathrm{E}}_2}{2.303\times 8.314}\left[\frac{10}{298\times 308}\right]$
log2=E22.303×8.314[10298×308]$\log 2=\frac{{\mathrm{E}}_2}{2.303\times 8.314}\left[\frac{10}{298\times 308}\right]$
Ea=2.303×8.314×298×308×log210$E_a=\frac{2.303\times 8.314\times 298\times 308\times \log 2}{10}$
= 52897.78 Jmol-1
= 52.9 Jmol-1

---

9. The activation energy for the reaction 2HI(g) →$\to$H2(g) + I2(g) is 209.5 kJmol-1 at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
Ans. In the given case:
Ea = 209.5 kJmol-1 = 209500 J mol-1
T = 581 K
R = 8.314 JK-1mol-1
Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:
x=e−Ea/RT$\mathrm{x}={\mathrm{e}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{R}\mathrm{T}}$
lnx=−Ea/RT$\ln x=-E_a/RT$
logx=−Ea2303RT$\log x=-\frac{E_a}{2303RT}$
logx=209500Jmol−12.303×8.314JK−1mol−1×581=18.8323$\log x=\frac{209500{\mathrm{J}\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}}{2.303\times 8.314{\mathrm{J}\mathrm{K}}^{-1}{\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}\times 581}=18.8323$
Now, x = Anti log (18.8323)
= Anti log 19¯¯¯¯¯.1677$\overline{19}.1677$
=1.471×10−19$=1.471\times {10}^{-19}$

Chapter End question Part-1

---

1. From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
(i) 3NO(g)→$\to$N2O(g); Rate = k[NO]2
(ii) H2O2(aq) + 3I-(aq) + H+ →$\to$ 2H2O(l) + I−3$I_3^{-}$; Rate = k[H2O][I-]
(iii) CH3CHO(g)→$\to$CH4(g) + CO(g) ;Rate= k[CH3CHO]3/2
(iv) C2H5Cl(g) →$\to$C2H2(g) + HCl(g); Rate =k[C2H5Cl]
Ans. (i) Given rate = k[NO]2
Therefore, order of the reaction = 2
Dimension of k= Rate [NO]2$\mathrm{k}=\frac{\text{ Rate }}{[\mathrm{N}\mathrm{O}{]}^2}$
=mo1L−1s−1(mo1L−1)2$=\frac{mo1L^{-1}{s}^{-1}}{{\left(mo1L^{-1}\right)}^2}$
=molL−1s−1mol2L−2$=\frac{\mathrm{mol}{L}^{-1}{s}^{-1}}{mo{l}^2{L}^{-2}}$
= L mol-1s-1
(ii) Given rate = k[H2O][I-]
Therefore, order of the reaction = 2
Dimension of k= Rate [H2O2][I−]$\mathrm{k}=\frac{\text{ Rate }}{\left[{\mathrm{H}}_2{\mathrm{O}}_2\right]\left[{\mathrm{I}}^{-}\right]}$
=molL−1s−1(molL−1)(molL−1)$=\frac{{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{L}}^{-1}{\mathrm{s}}^{-1}}{\left(\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}\right)\left(\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}\right)}$
= L mol-1s-1
(iii) Given rate = k[CH3CHO]3/2
Therefore, order of reaction =32$\frac{3}{2}$
Dimension of k= Rate [CH3CHO]32$\mathrm{k}=\frac{\text{ Rate }}{{\left[{\mathrm{C}\mathrm{H}}_3\mathrm{C}\mathrm{H}\mathrm{O}\right]}^{\frac{3}{2}}}$
=molL−1s−1(molL−1)32$=\frac{\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}}{{\left({\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{L}}^{-1}\right)}^{\frac{3}{2}}}$
=molL−1s−1mol32L−32$=\frac{mol{L}^{-1}{s}^{-1}}{mo{l}^{\frac{3}{2}}{L}^{-\frac{3}{2}}}$
=L12mol−12s−1$={\mathrm{L}}^{\frac{1}{2}}{\mathrm{m}\mathrm{o}\mathrm{l}}^{-\frac{1}{2}}{\mathrm{s}}^{-1}$
(iv) Given rate = k[C2H5Cl]
Therefore, order of the reaction = 1
Dimension of k= Rate [C2H5Cl]$\mathrm{k}=\frac{\text{ Rate }}{\left[{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{C}\mathrm{l}\right]}$
=molL−1s−1molL−1$=\frac{\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}}{\mathrm{mol}{\mathrm{L}}^{-1}}$
= s-1

---

2. For the reaction: 2A + B →$\to$A2B the rate = k[A][B]2
with k = 2.0 ×$\times$ 10-6 mol-2L2s-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L-1, [B] = 0.2 mol L-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L-1.
Ans. The initial rate of the reaction is
Rate = k[A][B]2
=(2.0×10−5mol−2L2s−1)$=\left(2.0\times {10}^{-5}{\mathrm{m}\mathrm{o}\mathrm{l}}^{-2}{\mathrm{L}}^2{\mathrm{s}}^{-1}\right)$(0.1 mol L-1)(0.2 mol L-1)
=8.0×10−9mol−2L2s−1$=8.0\times {10}^{-9}{\mathrm{m}\mathrm{o}\mathrm{l}}^{-2}{\mathrm{L}}^2{\mathrm{s}}^{-1}$
When [A] is reduced from 0.1 mol L-1 to 0.06 mol L-1, the concentration of A reacted =
(0.1 - 0.06) mol L-1 = 0.004 mol L-1
Therefore, concentration of B reacted =12×0.04molL−1=0.02molL−1$=\frac{1}{2}\times 0.04\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}=0.02\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}$
Then, concentration of B available, [B] = (0.2 - 0.02) mol L-1 = 0.18 mol L-1
After [A] is reduced to 0.06 mol L-1, the rate of the reaction is given by,
Rate = k[A][B]2
= (2.0 ×$\times$10-6 mol-2L2s-1)(0.06 mol L-1)(0.18 mol L-1)
= 3.98 mol L-1s-1

---

3. The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 ×$\times$ 10-4 mol-1Ls-1?
Ans. The decomposition of NH3 on platinum surface is represented by the following equation.
2NH3(g)⟶PtN2(g)+3H2(g)$2{\mathrm{N}\mathrm{H}}_{3(\mathrm{g})}\stackrel{\mathrm{P}\mathrm{t}}{⟶}{\mathrm{N}}_{2(\mathrm{g})}+3{\mathrm{H}}_{2(\mathrm{g})}$

Therefore, Rate = −12d[NH3]dt=d[N2]dt=13d[H2]dt$-\frac{1}{2}\frac{\mathrm{d}\left[{\mathrm{N}\mathrm{H}}_3\right]}{\mathrm{d}\mathrm{t}}=\frac{\mathrm{d}\left[{\mathrm{N}}_2\right]}{\mathrm{d}\mathrm{t}}=\frac{1}{3}\frac{\mathrm{d}\left[{\mathrm{H}}_2\right]}{\mathrm{d}\mathrm{t}}$
However, it is given that the reaction is of zero order.
Therefore,
−12d[NH3]dt=d[N2]dt=13d[H2]dt$-\frac{1}{2}\frac{\mathrm{d}\left[{\mathrm{N}\mathrm{H}}_3\right]}{\mathrm{d}\mathrm{t}}=\frac{\mathrm{d}\left[{\mathrm{N}}_2\right]}{\mathrm{d}\mathrm{t}}=\frac{1}{3}\frac{\mathrm{d}\left[{\mathrm{H}}_2\right]}{\mathrm{d}\mathrm{t}}$= k
= 2.5 ×$\times$ 10-4 mol-1Ls-1
Therefore, the rate of production of N2 is
d[N2]dt=2.5×10−4molL−1s−1$\frac{\mathrm{d}\left[{\mathrm{N}}_2\right]}{\mathrm{d}\mathrm{t}}=2.5\times {10}^{-4}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$
And, the rate of production of H2 is
d[H2]dt=3×2.5×10−4molL−1s−1$\frac{\mathrm{d}\left[{\mathrm{H}}_2\right]}{\mathrm{d}\mathrm{t}}=3\times 2.5\times {10}^{-4}\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$
= 7.5 ×$\times$ 10-4 mol-1Ls-1

---

4. The decomposition of dimethyl ether leads to the formation of CH4,H2 and CO and the reaction rate is given by Rate = k[CH3OCH3]3/2. The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., Rate = k(PCH3OCH3)3/2${\left({\mathrm{P}}_{\mathrm{C}{\mathrm{H}}_3\mathrm{O}{\mathrm{C}\mathrm{H}}_3}\right)}^{3/2}$. If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?
Ans. If pressure is measured in bar and time in minutes, then
Unit of rate = bar min-1
Rate = k(PCH3OCH3)3/2${\left({\mathrm{P}}_{\mathrm{C}{\mathrm{H}}_3\mathrm{O}{\mathrm{C}\mathrm{H}}_3}\right)}^{3/2}$
k = Rate(PCH3OCH3)3/2$\frac{Rate}{{\left({\mathrm{P}}_{\mathrm{C}{\mathrm{H}}_3\mathrm{O}{\mathrm{C}\mathrm{H}}_3}\right)}^{3/2}}$
Therefore, unit of rate constants (k)= bar min−1bar3/2$(\mathrm{k})=\frac{\text{ bar }\stackrel{-1}{min}}{{\mathrm{b}\mathrm{a}\mathrm{r}}^{3/2}}$
=bar-1/2 min-1

---

5. Mention the factors that affect the rate of a chemical reaction.
Ans. The factors that affect the rate of a reaction areas follows.
(i) Concentration of reactants (pressure in case of gases)
(ii) Temperature
(iii) Presence of a catalyst

---

6. A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is
(i) doubled
(ii) reduced to half?
Ans. Let the concentration of the reactant be [A] = a
Rate of reaction, k[A]2= ka2
(i) If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be
R' = k(2a)2
= 4ka2
= 4 R
Therefore, the rate of the reaction would increase by 4 times.
(ii) If the concentration of the reactant is reduced to half, i.e. [A]=12a$[\mathrm{A}]=\frac{1}{2}\mathrm{a}$, then the rate of the reaction would be R′=k(12a)2${\mathrm{R}}^{\mathrm{\prime }}=\mathrm{k}{\left(\frac{1}{2}\mathrm{a}\right)}^2$
=14ka2$=\frac{1}{4}{\mathrm{k}\mathrm{a}}^2$
=14R$=\frac{1}{4}R$
Therefore, the rate of the reaction would be reduced to One - fourth.

---

7. What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?
Ans. The rate constant is nearly doubled with a rise in temperature by 100C for a chemical reaction.
The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation,
k=Ae−Ea/RT$\mathrm{k}={\mathrm{A}\mathrm{e}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{R}\mathrm{T}}$
where, k is the rate constant,
A is the Arrhenius factor or the frequency factor,
R is the gas constant,
T is the temperature, and
Ea is the energy of activation for the reaction.

---

8. In a pseudo first order hydrolysis of ester in water, the following results were obtained:

t/s	0	30	60	90
[Ester]mol L-1	0.55	0.31	0.17	0.085

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
Ans. (i) Average rate of reaction between the time interval, 30 to 60 seconds,=d[Ester]dt$=\frac{\mathrm{d}[\mathrm{E}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{r}]}{\mathrm{d}\mathrm{t}}$
=0.31−0.1760−30$=\frac{0.31-0.17}{60-30}$
=0.1430$=\frac{0.14}{30}$
= 4.67 ×$\times$ 10-3 mol L-1s-1
(ii) For a pseudo first order reaction,
k=2.303tlog[R]0[R]$\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{[\mathrm{R}{]}_0}{[\mathrm{R}]}$
For t= 30 s, k1=2.30330log0.550.31${\mathrm{k}}_1=\frac{2.303}{30}\log \frac{0.55}{0.31}$
= 1.911 ×$\times$ 10-2 s-1
For t= 60 s, k2=2.30360log0.550.17${\mathrm{k}}_2=\frac{2.303}{60}\log \frac{0.55}{0.17}$
= 1.957 ×$\times$ 10-2 s-1
For t= 90 s,
k3=2.30390log0.550.085${\mathrm{k}}_3=\frac{2.303}{90}\log \frac{0.55}{0.085}$
= 2.075 ×$\times$ 10-2 s-1
Then, average rate constant, k=k1+k2+k33$\mathrm{k}=\frac{{\mathrm{k}}_1+{\mathrm{k}}_2+{\mathrm{k}}_3}{3}$
=(1.911×10−2)+(1.957×10−2)+(2.075×10−2)3$=\frac{\left(1.911\times {10}^{-2}\right)+\left(1.957\times {10}^{-2}\right)+\left(2.075\times {10}^{-2}\right)}{3}$
= 1.98 ×$\times$ 10-2 s-1

---

9. A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
Ans. (i) The differential rate equation will be
−d[R]dt=k[A][B]2$-\frac{\mathrm{d}[\mathrm{R}]}{\mathrm{d}\mathrm{t}}=\mathrm{k}[\mathrm{A}][\mathrm{B}{]}^2$
(ii) If the concentration of B is increased three times, then
−d[R]dt=k[A][3B]2$-\frac{\mathrm{d}[\mathrm{R}]}{\mathrm{d}\mathrm{t}}=\mathrm{k}[\mathrm{A}][3\mathrm{B}{]}^2$
= 9.k[A][B]2
Therefore, the rate of reaction will increase 9 times.
(iii) When the concentrations of both A and B are doubled,
−d[R]dt=k[A][B]2$-\frac{\mathrm{d}[\mathrm{R}]}{\mathrm{d}\mathrm{t}}=\mathrm{k}[\mathrm{A}][\mathrm{B}{]}^2$
= k[2A][2B]2
= 8.k[A][B]2
Therefore, the rate of reaction will increase 8 times.

---

10. In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:

A/mol L-1	0.20	0.20	0.40
B/mol L-1	0.30	0.10	0.05
r0/mol L-1s-1	5.07×$\times$ 10-5	5.07×$\times$ 10-5	1.43×$\times$ 10-4

What is the order of the reaction with respect to A and B?
Ans. Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore, r0 = k[A]x[B]y
5.07×$\times$ 10-5 = k[0.20]x[0.30]y.......(i)
5.07×$\times$ 10-5 = k[0.20]x[0.10]y.......(i)
1.43×$\times$ 10-4 = k[0.40]x[0.05]y.......(i)
Dividing equation (i) by (ii), we obtain
5.07×10−55.07×10−5=k[0.20]x[0.30]yk[0.20]2[0.10]y$\frac{5.07\times {10}^{-5}}{5.07\times {10}^{-5}}=\frac{k[0.20{]}^x[0.30{]}^y}{k[0.20{]}^2[0.10{]}^y}$
1=[0.30]y[0.10]y$1=\frac{[0.30{]}^y}{[0.10{]}^y}$
(0.300.10)0=(0.300.10)y${\left(\frac{0.30}{0.10}\right)}^0={\left(\frac{0.30}{0.10}\right)}^y$
y = 0
Dividing equation (iii) by (ii), we obtain
1.43×10−45.07×10−5=k[0.40]x[0.05]yk[0.20]x[0.30]y$\frac{1.43\times {10}^{-4}}{5.07\times {10}^{-5}}=\frac{k[0.40{]}^x[0.05{]}^y}{k[0.20{]}^x[0.30{]}^y}$
1.43×10−45.07×10−5=[0.40]y[0.20]y$\frac{1.43\times {10}^{-4}}{5.07\times {10}^{-5}}=\frac{[0.40{]}^y}{[0.20{]}^y}$
2.821 = 2x
log 2.821 = x log 2(Taking log on both sides)
x=log2.821log2$\mathrm{x}=\frac{\log 2.821}{\log 2}$
= 1.496
= 1.5 (approximately)
Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.

---

11. The following results have been obtained during the kinetic studies of the reaction:
2A + B →$\to$ C + D

Experiment	A/mol L-1	B/mol L-1	Initial rate/mol L-1min-1
I	0.1	0.1	6.0×$\times$10-3
II	0.3	0.2	7.2 ×$\times$ 10-2
III	0.3	0.4	2.88 ×$\times$ 10-1
IV	0.4	0.1	2.4×$\times$10-2

Determine the rate law and the rate constant for the reaction.
Ans. Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore, rate of the reaction is given by,
Rate = k[A]x[B]y
According to the question,
6.0 ×$\times$ 10-3 = k[0.1]x[0.1]y.......(i)
7.2 ×$\times$ 10-2 = k[0.3]x[0.2]y.......(ii)
2.88 ×$\times$ 10-1 = k[0.3]x[0.4]y.......(iii)
2.40 ×$\times$ 10-2 = k[0.4]x[0.1]y........(iv)
Dividing equation (iv) by (i), we obtain
2.40×10−26.0×10−3=k[0.4]x[0.1]yk[0.1]x[0.1]y$\frac{2.40\times {10}^{-2}}{6.0\times {10}^{-3}}=\frac{\mathrm{k}[0.4{]}^{\mathrm{x}}[0.1{]}^y}{\mathrm{k}[0.1{]}^{\mathrm{x}}[0.1{]}^y}$
4=[0.4]π[0.1]π$4=\frac{[0.4{]}^{\pi }}{[0.1{]}^{\pi }}$
4=(0.40.1)x$4={\left(\frac{0.4}{0.1}\right)}^x$
(4)1 = 4x
x = 1
Dividing equation (iii) by (ii), we obtain
2.88×10−17.2×10−2=k[0.3]x[0.4]xk[0.3]x[0.2]x$\frac{2.88\times {10}^{-1}}{7.2\times {10}^{-2}}=\frac{\mathrm{k}[0.3{]}^{\mathrm{x}}[0.4{]}^{\mathrm{x}}}{\mathrm{k}[0.3{]}^{\mathrm{x}}[0.2{]}^{\mathrm{x}}}$
4=(0.40.2)y$4={\left(\frac{0.4}{0.2}\right)}^y$
4 = 2y
22 = 2y
y = 2
Therefore, the rate law is
Rate = k[A][B]2
k= Rate [A][B]2$\mathrm{k}=\frac{\text{ Rate }}{[\mathrm{A}][\mathrm{B}{]}^2}$
From experiment I, we obtain
k=6.0×10−3molL−1min−1(0.1molL−1)(0.1molL−1)2$\mathrm{k}=\frac{6.0\times {10}^{-3}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{L}}^{-1}{\mathrm{m}\mathrm{i}\mathrm{n}}^{-1}}{\left(0.1\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}\right){\left(0.1\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}\right)}^2}$
= 6.0 L2mol-2min-1
From experiment II, we obtain,
k=7.2×10−2molL−1min−1(0.3molL−1)(0.2molL−1)2$\mathrm{k}=\frac{7.2\times {10}^{-2}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{L}}^{-1}{\mathrm{m}\mathrm{i}\mathrm{n}}^{-1}}{\left(0.3\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}\right){\left(0.2\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}\right)}^2}$
= 6.0 L2mol-2min-1
From experiment III, we obtain
k=2.88×10−1molL−1min−1(0.3molL−1)(0.4molL−1)2$\mathrm{k}=\frac{2.88\times {10}^{-1}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{L}}^{-1}{\mathrm{m}\mathrm{i}\mathrm{n}}^{-1}}{\left(0.3\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}\right){\left(0.4\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}\right)}^2}$
= 6.0 L2mol-2min-1
From experiment IV, we obtain
k=2.40×10−2molL−1min−1(0.4molL−1)(0.1molL−1)2$\mathrm{k}=\frac{2.40\times {10}^{-2}\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}{\mathrm{m}\mathrm{i}\mathrm{n}}^{-1}}{\left(0.4{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{L}}^{-1}\right){\left(0.1\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}\right)}^2}$
= 6.0 L2mol-2min-1
Therefore, rate constant, k = 6.0 L2mol-2min-1

---

12. The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:

Experiment	A/mol L-1	B/mol L-1	Initial rate/mol L-1min-1
I	0.1	0.1	2.0×$\times$10-2
II	-	0.2	4.0×$\times$10-2
III	0.4	0.4	-
IV	-	0.2	2.0×$\times$10-2

Ans. The given reaction is of the first order with respect to A and of zero order with respect to B.
Therefore, the rate of the reaction is given by,
Rate = k [A]1[B]0
Rate = k [A]
From experiment I, we obtain
2.0 ×$\times$ 10-2 mol L-1min-1 =k(0.1 mol L-1)
k = 0.2 min-1
From experiment II, we obtain
4.0 ×$\times$ 10-2 mol L-1min-1 = 0.2 min-1[A]
[A] = 0.2 mol L-1
From experiment III, we obtain
Rate = 0.2 min-1×$\times$ 0.4 mol L-1
= 0.08 mol L-1min-1
From experiment IV, we obtain
2.0 ×$\times$ 10-2 mol L-1min-1 = 0.2 min-1[A]
[A] = 0.1 mol L-1

---

13. Calculate the half-life of a first order reaction from their rate constants given below:
(i) 200 s-1
(ii) 2 min-1
(iii) 4 year-1
Ans. (i) Half-life, t1/2=0.693k${\mathrm{t}}_{1/2}=\frac{0.693}{\mathrm{k}}$
=0.693200min−1$=\frac{0.693}{200\stackrel{-1}{min}}$
= 3.4 ×$\times$10-3 s(approximately)
(ii) Half-life, t1/2=0.693k${\mathrm{t}}_{1/2}=\frac{0.693}{\mathrm{k}}$
=0.6932min−1$=\frac{0.693}{2{\mathrm{m}\mathrm{i}\mathrm{n}}^{-1}}$
= 0.35 min (approximately)
(iii) Half-life, t1/2=0.693k$t_{1/2}=\frac{0.693}{\mathrm{k}}$
=0.6934years−1$=\frac{0.693}{4{\mathrm{y}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{s}}^{-1}}$
= 0.173 years (approximately)

---

14. The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.
Ans. Here, k=0.693t1/2$\mathrm{k}=\frac{0.693}{{\mathrm{t}}_{1/2}}$
=0.6935730 years −1$=\frac{0.693}{5730}{\text{ years }}^{-1}$
It is known that,
t=2.303klog[R]0[R]$\mathrm{t}=\frac{2.303}{\mathrm{k}}\log \frac{[\mathrm{R}{]}_0}{[\mathrm{R}]}$
=2.3030.6935730log10080$=\frac{2.303}{\frac{0.693}{5730}}\log \frac{100}{80}$
= 1845 years (approximately)
Hence, the age of the sample is 1845 years.

---

15. The experimental data for decomposition of N2O5[2N2O5 →$\to$4NO2 + O2] in gas phase at 318 K are given below:

t(s)	0	400	800	1200	1600	2000	2400	2800	3200
102×$\times$[N2O5] mol L-1	1.63	1.36	1.14	0.93	0.78	0.64	0.53	0.43	0.35

(i) Plot [N2O5] against t.
(ii) Find the half-life period for the reaction.
(iii) Draw a graph between log[N2O5] and t.
(iv) What is the rate law?
(v) Calculate the rate constant.
(vi) Calculate the half-life period from k and compare it with (ii).
Ans.

(ii) Time corresponding to the concentration, 1.630×1022$\frac{1.630\times {10}^2}{2}$mol L-1 = 81.5 mol L-1 is the half-life. From the graph, the half-life is obtained as 1450 s.
(iii)

t(s)	102×$\times$[N2O5]/ mol L-1	log[N2O5]
0	1.63	-1.79
400	1.36	-1.87
800	1.14	-1.94
1200	0.93	-2.03
1600	0.78	-2.11
2000	0.64	-2.19
2400	0.53	-2.28
2800	0.43	-2.37
3200	0.35	-2.46

(iv) The given reaction is of the first order as the plot, log[N2O5] v/s t, is a straight line. Therefore, the rate law of the reaction is Rate = k[N2O5]

(v) From the plot, log[N2O5] v/s t, we obtain
Slope = −2.46−(−1.79)3200−0$\frac{-2.46-(-1.79)}{3200-0}$
=−0.673200$=\frac{-0.67}{3200}$
Again, slope of the line of the plot log[N2O5] v/s t is given by−k2.303$-\frac{\mathrm{k}}{2.303}$. Therefore, we obtain,−k2.303=−0.673200$-\frac{\mathrm{k}}{2.303}=-\frac{0.67}{3200}$

Chapter End question Part-2

---

16. The rate constant for a first order reaction is 60 s-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?
Ans. It is known that,
t=2.303klog[R]0[R]$t=\frac{2.303}{k}\log \frac{[R{]}_0}{[R]}$
=2.30360s−1log11/16$=\frac{2.303}{60s^{-1}}\log \frac{1}{1/16}$
=2.30360s−1log16$=\frac{2.303}{60s^{-1}}\log 16$
=4.6×10−2s$=4.6\times {10}^{-2}\mathrm{s}$(approximately)
Hence, the required time is 4.6×10−2s$4.6\times {10}^{-2}\mathrm{s}$.

---

17. During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1μ$\mu$g of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.
Ans. Here, k=0.693t1/2=0.69328.1y−1$\mathrm{k}=\frac{0.693}{{\mathrm{t}}_{1/2}}=\frac{0.693}{28.1}{\mathrm{y}}^{-1}$
It is known that, t=2.303klog[R]0[R]$\mathrm{t}=\frac{2.303}{\mathrm{k}}\log \frac{[\mathrm{R}{]}_0}{[\mathrm{R}]}$
10=2.3030.69328.1log1[R]$10=\frac{2.303}{\frac{0.693}{28.1}}\log \frac{1}{[\mathrm{R}]}$
10=2.3030.6930.69328.1(−log[R])$10=\frac{\frac{2.303}{0.693}}{\frac{0.693}{28.1}}(-\log [\mathrm{R}])$
log[R]=−10×0.6932.303×28.1$\log [\mathrm{R}]=-\frac{10\times 0.693}{2.303\times 28.1}$
[R] = anitlog(-0.1071)
= antilog(1¯¯¯.8929$\overline{1}.8929$)
= 0.7814 μ$\mu$g
Therefore, = 0.7814 μ$\mu$g of 90Sr will remain after 10 years.
Again, t=2.303klog[R]0[R]$t=\frac{2.303}{k}\log \frac{[\mathrm{R}{]}_0}{[R]}$
60=2.3030.69328.1log1[R]$60=\frac{2.303}{\frac{0.693}{28.1}}\log \frac{1}{[\mathrm{R}]}$
log[R]=60×0.6932.303×28.1$\log [\mathrm{R}]=\frac{60\times 0.693}{2.303\times 28.1}$
[R] = antilog(-0.6425)
=antilog(1¯¯¯.3575)$=\mathrm{anti}\log (\overline{1}.3575)$
= 0.2278 μ$\mu$g
Therefore, 0.2278 μ$\mu$g of 90Sr will remain after 60 years.

---

18. For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Ans. For a first order reaction, the time required for 99% completion is
t1=2.303klog100100−99$t_1=\frac{2.303}{k}\log \frac{100}{100-99}$
=2.303klog100$=\frac{2.303}{\mathrm{k}}\log 100$
=2×2.303k$=2\times \frac{2.303}{\mathrm{k}}$
For a first order reaction, the time required for 90% completion is
t2=2.303klog100100−90${\mathrm{t}}_2=\frac{2.303}{\mathrm{k}}\log \frac{100}{100-90}$
=2.303klog10$=\frac{2.303}{\mathrm{k}}\log 10$
=2.303k$=\frac{2.303}{\mathrm{k}}$
Therefore, t1 = 2t2
Hence, the time required for 99% completion of a first order reaction is twice the time required for the completion of 90% of the reaction.

---

19. A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.
Ans. For a first order reaction,
t1=2.303klog[R]0[R]${\mathrm{t}}_1=\frac{2.303}{\mathrm{k}}\log \frac{[\mathrm{R}{]}_0}{[\mathrm{R}]}$
k=2.30340minlog100100−30$\mathrm{k}=\frac{2.303}{40\mathrm{m}\mathrm{i}\mathrm{n}}\log \frac{100}{100-30}$
=2.30340minlog107$=\frac{2.303}{40\mathrm{m}\mathrm{i}\mathrm{n}}\log \frac{10}{7}$
= 8.918 ×$\times$ 10-3min-1
Therefore, t1/2 of the decomposition reaction is
t1/2=0.693k$t_{1/2}=\frac{0.693}{\mathrm{k}}$
=0.6938.918×10−3min$=\frac{0.693}{8.918\times {10}^{-3}}\mathrm{m}\mathrm{i}\mathrm{n}$
= 77.7 min (approximately)

---

20. For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

t(sec)	P(mm of Hg)
0	35.0
360	54.0
720	63.0

Calculate the rate constant.
Ans. The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.

After time, t, total pressure, = (P0 - p) + p + p
Pt = P0 + p
p = Pt - P0
Therefore, 

For a first order reaction,
k=2.303tlogP0P0−p$\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{P}}_0}{{\mathrm{P}}_0-\mathrm{p}}$
=2.303tlogP02P0−Pt$=\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{P}}_0}{2{\mathrm{P}}_0-{\mathrm{P}}_{\mathrm{t}}}$
When t = 360 s, k=2.303360slog35.02×35.0−54.0$\mathrm{k}=\frac{2.303}{360\mathrm{s}}\log \frac{35.0}{2\times 35.0-54.0}$
=2.175×10−3s−1$=2.175\times {10}^{-3}{\mathrm{s}}^{-1}$
When t = 720 s, k=2.303720slog35.02×35.0−63.0$\mathrm{k}=\frac{2.303}{720\mathrm{s}}\log \frac{35.0}{2\times 35.0-63.0}$
=2.235×10−3s−1$=2.235\times {10}^{-3}{\mathrm{s}}^{-1}$
Hence, the average value of rate constant is
k=(2.175×10−3)+(2.235×10−3)2s−1$\mathrm{k}=\frac{\left(2.175\times {10}^{-3}\right)+\left(2.235\times {10}^{-3}\right)}{2}{\mathrm{s}}^{-1}$
=2.21×10−3s−1$=2.21\times {10}^{-3}{\mathrm{s}}^{-1}$

---

21. The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.
SO2Cl2(g)→SO2(g) + Cl2(g)$\text{S}{\text{O}}_{\text{2}}\text{C}{\text{l}}_{\text{2}}\text{(g)}\to \text{S}{\text{O}}_{\text{2}}\text{(g)}\text{ + }\text{C}{\text{l}}_{\text{2}}\text{(g)}$

Experiment	Time/s-1	Total pressure/atm
1	0	0.5
2	100	0.6

Calculate the rate of the reaction when total pressure is 0.65 atm.
Ans. The thermal decomposition of SO2Cl2 at a constant volume is represented by the following equation.

t=0t=tSO2Cl2(g)P0P0−p→SO2(g)0p+Cl2(g)0p$\underset{t=t}{\underset{t=0}{}}\underset{P_0-p}{\underset{P_0}{\text{S}{\text{O}}_2\text{C}{\text{l}}_{2(\text{g})}}}\to \underset{p}{\underset{0}{\text{S}{{\text{O}}_{}}_{2(\text{g})}}}+\underset{p}{\underset{0}{\text{C}{\text{l}}_{2(\text{g})}}}$
After time, t, total pressure, Pt = (P0 - p) + p + p
Pt = P0 + p
p = Pt - P0
Therefore,
P0 - p = P0 - (Pt - P0)
= 2P0 - Pt
For a first order reaction,
k=2.303tlogP0P0−p$\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{P}}_0}{{\mathrm{P}}_0-\mathrm{p}}$
=2.303tlogP02P0−Pt$=\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{P}}_0}{2{\mathrm{P}}_0-{\mathrm{P}}_{\mathrm{t}}}$
When t= 100 s, k=2.303100slog0.52×0.5−0.6$\mathrm{k}=\frac{2.303}{100\mathrm{s}}\log \frac{0.5}{2\times 0.5-0.6}$
=2.231×10−3s−1$=2.231\times {10}^{-3}{\mathrm{s}}^{-1}$
When Pt = 0.65 atm,
P0 +p = 0.65
p = 0.65 - P0
= 0.65 – 0.5
= 0.15 atm
Therefore, when the total pressure is 0.65 atm, pressure of SOCl2 is
PSOCl2=P0−p${\mathrm{P}}_{{\mathrm{S}\mathrm{O}\mathrm{C}\mathrm{l}}_2}={\mathrm{P}}_0-\mathrm{p}$
= 0.5 – 0.15
= 0.35 atm
Therefore, the rate of equation, when total pressure is 0.65 atm, is given by,
Rate = k(PSoCl2)$\mathrm{k}\left({\mathrm{P}}_{{\mathrm{S}\mathrm{o}\mathrm{C}\mathrm{l}}_2}\right)$
= (2.23×10−3s−1)(0.35atm)$\left(2.23\times {10}^{-3}{s}^{-1}\right)(0.35atm)$
=7.8×10−4atms−1$=7.8\times {10}^{-4}\mathrm{a}\mathrm{t}\mathrm{m}{\mathrm{s}}^{-1}$

---

22. The rate constant for the decomposition of N2O5 at various temperatures is given below:

T/°C	0	20	40	60	60
105 ×$\times$ k/s-1	0.0787	1.70	25.7	178	2140

Draw a graph between ln k and 1/T and calculate the values of A and Ea. Predict the rate constant at 30° and 50°C.
Ans. From the given data, we obtain

T/°C	0	20	40	60	60
T/K	273	293	313	333	353
1T/K−1$\frac{1}{\mathrm{T}}/{\mathrm{K}}^{-1}$	3.66 ×$\times$ 10-3	3.41` ×$\times$ 10-3	3.19 ×$\times$ 10-3	3.0 ×$\times$ 10-3	2.83 ×$\times$ 10-3
105 ×$\times$ k/s-1	0.0787	1.70	25.7	178	2140
ln K	-7.147	-4.075	-1.359	-0.577	3.063

Slope of the line,
y2−y1x2−x1=−12.301K$\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}=-12.301\mathrm{K}$
According to Arrhenius equation,
Slope = −EaR$-\frac{E_a}{R}$
Ea = -Slope ×$\times$ R
= -(-12.301 K) ×$\times$ (8.314 JK-1mol-1)
= 102.27 kJmol-1
Again,
lnk=lnA−EaRT$\ln \mathrm{k}=\ln \mathrm{A}-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}\mathrm{T}}$
lnA=lnk+EaRT$\ln A=\ln k+\frac{E_a}{RT}$
When T = 273 K
In k = –7.147
Then, In A = −7.147+102.27×1038.314×273$-7.147+\frac{102.27\times {10}^3}{8.314\times 273}$
=37.911
Therefore, A = 2.91 ×$\times$ 106
When, T = 30 + 273K = 303 K
1T=0.0033K=3.3×10−3K$\frac{1}{T}=0.0033\mathrm{K}=3.3\times {10}^{-3}\mathrm{K}$
Then, at 1T=3.3×10−3K$\frac{1}{\mathrm{T}}=3.3\times {10}^{-3}\mathrm{K}$,
In k = –2.8
Therefore, k = 6.08 ×$\times$ 10-2 s-1
Again, when T = 50 + 273K = 323K
1/T = 3.09 x 10-3
ln k= -0.6
k = 0.5488 s-1

---

23. The rate constant for the decomposition of hydrocarbons is 2.418 ×$\times$ 10-5 s-1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
Ans. k = 2.418 ×$\times$ 10-5 s-1
T= 546 K
Ea = 179.9 kJ mol-1
=179.9×103Jmol−1$=179.9\times {10}^3{\mathrm{J}\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}$
According to the Arrhenius equation,
k=Ae−Ea/RT$\mathrm{k}={\mathrm{A}\mathrm{e}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{R}\mathrm{T}}$
lnk=lnA−EzRT$\ln \mathrm{k}=\ln \mathrm{A}-\frac{{\mathrm{E}}_{\mathrm{z}}}{\mathrm{R}\mathrm{T}}$
logk=logA−Ea2.303RT$\log k=\log A-\frac{E_a}{2.303RT}$
logA=logk+Ea2.303RT$\log A=\log k+\frac{E_a}{2.303RT}$
=log(2.418×10−5s−1)+179.9×103Jmol−12.303×8.314Jk−1mol−1×546K$=\log \left(2.418\times {10}^{-5}{s}^{-1}\right)+\frac{179.9\times {10}^3{\mathrm{J}\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}}{2.303\times 8.314{\mathrm{J}\mathrm{k}}^{-1}{\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}\times 546\mathrm{K}}$
= (0.3835 - 5) + 17.2082
= 12.5917
Therefore, A = antilog (12.5917)
= 3.9 ×$\times$ 1012 s-1(approximately)

---

24. Consider a certain reaction A →$\to$ Products with k = 2.0 ×$\times$ 10-2 s-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L-1.
Ans. k = 2.0 ×$\times$ 10-2s-1
T= 100 s
[A]0 = 1.0 mol L-1
Since the unit of k is s-1, the given reaction is a first order reaction.
Therefore, k=2.303tlog[A]0[A]$\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{[\mathrm{A}{]}_0}{[\mathrm{A}]}$
2.0×10−2s−1=2.303100log1.0[A]$2.0\times {10}^{-2}{s}^{-1}=\frac{2.303}{100}\log \frac{1.0}{[\mathrm{A}]}$
2.0×10−2s−1=2.303100(−log[A])$2.0\times {10}^{-2}{s}^{-1}=\frac{2.303}{100}(-\log [A])$
−log[A]=2.0×10−2×1002.303$-\log [\mathrm{A}]=\frac{2.0\times {10}^{-2}\times 100}{2.303}$
[A]= antilog (−2.0×10−2×1002.303)$[\mathrm{A}]=\text{ antilog }\left(-\frac{2.0\times {10}^{-2}\times 100}{2.303}\right)$
= 0.135 mol L-1(approximately)
Hence, the remaining concentration of A is 0.135 mol L-1.

---

25. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
Ans. For a first order reaction,k=2.303tlog[R]0[R]$\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{[\mathrm{R}{]}_0}{[\mathrm{R}]}$
It is given that, t1/2 = 3.00 hours
Therefore, k=0.693t1/2$\mathrm{k}=\frac{0.693}{{\mathrm{t}}_{1/2}}$
=0.6933h−1$=\frac{0.693}{3}{\mathrm{h}}^{-1}$
= 0.231 h-1
Then,0.231h−1=2.3038hlog[R]0[R]$0.231{\mathrm{h}}^{-1}=\frac{2.303}{8\mathrm{h}}\log \frac{[\mathrm{R}{]}_0}{[\mathrm{R}]}$
log[R]0[R]=0.231h−1×8h2.303$\log \frac{[\mathrm{R}{]}_0}{[\mathrm{R}]}=\frac{0.231{\mathrm{h}}^{-1}\times 8\mathrm{h}}{2.303}$
[R]0[R]=antilog(0.8024)$\frac{[\mathrm{R}{]}_0}{[\mathrm{R}]}=\mathrm{antilog}(0.8024)$
[R]0[R]=6.3445$\frac{[\mathrm{R}{]}_0}{[\mathrm{R}]}=6.3445$
[R][R]0=0.1576$\frac{[\mathrm{R}]}{[\mathrm{R}{]}_0}=0.1576$(approx)
= 0.158
Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.

---

26. The decomposition of hydrocarbon follows the equation k = (4.5 ×$\times$ 1011 s-1)e-28000K/T.
Calculate Ea.
Ans. The given equation is k = (4.5 ×$\times$ 1011 s-1)e-28000K/T.……………..(i)
Arrhenius equation is given by,
k=Ae−Ea/RT$\mathrm{k}=\mathrm{A}{\mathrm{e}}^{-{\mathrm{E}}_a/\mathrm{R}\mathrm{T}}$ ……………….(ii)
From equation (i) and (ii), we obtain
EaRT=28000KT$\frac{E_a}{RT}=\frac{28000K}{T}$
Ea=R×28000K${\mathrm{E}}_{\mathrm{a}}=\mathrm{R}\times 28000\mathrm{K}$
= 8.314 J K-1mol-1 ×$\times$ 28000 K
= 232792 J mol-1
= 232.792 kJ mol-1

---

27. The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.34 - 1.25 ×$\times$ 104K/T. Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
Ans. Arrhenius equation is given by,
k=Ae−EzRT$\mathrm{k}=\mathrm{A}{\mathrm{e}}^{-{\mathrm{E}}_{\mathrm{z}}\mathrm{R}\mathrm{T}}$
lnk=lnA−EzRT$\ln \mathrm{k}=\ln \mathrm{A}-\frac{{\mathrm{E}}_{\mathrm{z}}}{\mathrm{R}\mathrm{T}}$
lnk=logA−EaRT$\ln \mathrm{k}=\log \mathrm{A}-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}\mathrm{T}}$
logk=logA−Ea2.303RT$\log \mathrm{k}=\log \mathrm{A}-\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{R}\mathrm{T}}$.........(i)
The given equation is
log k = 14.34 - 1.25 ×$\times$ 104K/T..........(ii)
From equation (i) and (ii), we obtain
Ez2.303RT=1.25×104KT$\frac{{\mathrm{E}}_{\mathrm{z}}}{2.303\mathrm{R}\mathrm{T}}=\frac{1.25\times {10}^4\mathrm{K}}{\mathrm{T}}$
Ea= =1.25×104K×2.303×R$=1.25\times {10}^4\mathrm{K}\times 2.303\times \mathrm{R}$
=1.25×104K×2.303×8.314JK−1mol$=1.25\times {10}^4\mathrm{K}\times 2.303\times 8.314{\mathrm{J}\mathrm{K}}^{-1}\mathrm{m}\mathrm{o}\mathrm{l}$
= 239339.3 J mol-1
= 239.34 kJ mol-1(approximately)
Also, when k1/2 = 256 minutes
k=0.693t1/2$\mathrm{k}=\frac{0.693}{{\mathrm{t}}_{1/2}}$
=0.693256$=\frac{0.693}{256}$
= 4.51 ×$\times$ 10-5 s-1
It is also given that, log K = 14.34 - 1.25 ×$\times$ 104 K/T
log(4.51 ×$\times$ 10-5) = 14.34 −1.25×104KT$-\frac{1.25\times {10}^4\mathrm{K}}{\mathrm{T}}$
log(0.654 - 5) = 14.34 −1.25×104KT$-\frac{1.25\times {10}^4\mathrm{K}}{\mathrm{T}}$
1.25×104KT=18.686$\frac{1.25\times {10}^4\mathrm{K}}{\mathrm{T}}=18.686$
T=1.25×104K18.686$\mathrm{T}=\frac{1.25\times {10}^4\mathrm{K}}{18.686}$
= 668.95 K
= 669 K (approximately)

---

28. The decomposition of A into product has value of k as 4.5 ×$\times$ 103 s-1 at 10°C and energy of activation 60 kJ mol-1. At what temperature would k be 1.5 ×$\times$ 104 s-1?
Ans. From Arrhenius equation, we obtain logk2k1=Ez2.303R(T2−T1T1T2)$\log \frac{k_2}{k_1}=\frac{E_z}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$
Also,k1 = 4.5 ×$\times$ 103 s-1
T1 =273 + 10 = 283 K
k2 = 1.5 ×$\times$ 104 s-1
Ea = 60 kJ mol-1 = 6.0 ×$\times$ 104 J mol-1
Then, log1.5×1044.5×103=6.0×104Jmol−12.303×8.314JK−1mol−1(T2−283283T2)$\log \frac{1.5\times {10}^4}{4.5\times {10}^3}=\frac{6.0\times {10}^4{\mathrm{J}\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}}{2.303\times 8.314{\mathrm{J}\mathrm{K}}^{-1}{\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}}\left(\frac{{\mathrm{T}}_2-283}{283{\mathrm{T}}_2}\right)$
0.5229=3133.6279(T2−283283T2)$0.5229=3133.6279\left(\frac{{\mathrm{T}}_2-283}{283{\mathrm{T}}_2}\right)$
0.5229×283T23133.627=T2−283$\frac{0.5229\times 283{\mathrm{T}}_2}{3133.627}={\mathrm{T}}_2-283$
0.0472 T2 = T2 - 283
0.9528 T2 = 283
T2 = 297.019 K(approximately)
= 297 K = 24°C
Hence, k would be 1.5 ×$\times$ 104 s-1 at 24°C.

---

29. The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is = 4 ×$\times$ 1010 s-1. Calculate k at 318 K and Ea.
Ans. For a first order reaction, t=2.303klogaa−x$\mathrm{t}=\frac{2.303}{\mathrm{k}}\log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}$
At 298 K, t=2.303klog10090$\mathrm{t}=\frac{2.303}{\mathrm{k}}\log \frac{100}{90}$
0.1054k$\frac{0.1054}{\mathrm{k}}$
At 308 K,
t′=2.303k′log10075${\mathrm{t}}^{\mathrm{\prime }}=\frac{2.303}{{\mathrm{k}}^{\mathrm{\prime }}}\log \frac{100}{75}$
=2.2877k′$=\frac{2.2877}{{\mathrm{k}}^{\mathrm{\prime }}}$
According to the question, t = t’
0.1054k=0.2877k′$\frac{0.1054}{\mathrm{k}}=\frac{0.2877}{{\mathrm{k}}^{\mathrm{\prime }}}$
k′k=2.7296$\frac{{\mathrm{k}}^{\mathrm{\prime }}}{\mathrm{k}}=2.7296$
From Arrhenius equation, we obtain
logk′k=Ea2.303R(T′−TTT′)$\log \frac{k^{\mathrm{\prime }}}{k}=\frac{E_a}{2.303R}\left(\frac{T^{\mathrm{\prime }}-T}{T{T}^{\mathrm{\prime }}}\right)$
log(2.7296)=Ea2.303×8.314(308−298298×308)$\log (2.7296)=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times 8.314}\left(\frac{308-298}{298\times 308}\right)$
Ez=2.303×8.314×298×308308−298${\mathrm{E}}_{\mathrm{z}}=\frac{2.303\times 8.314\times 298\times 308}{308-298}$
= 76640.096 J mol-1
= 76.64 kJ mol-1
To calculate k at 318 K,
It is given that, A = 4 ×$\times$ 1010 s-1, T = 318 K
Again, from Arrhenius equation, we obtain
logk=logA−Ez2.303RT$\log k=\log A-\frac{E_z}{2.303RT}$
=log(4×1010)−76.64×1032.303×8.314×318$=\log \left(4\times {10}^{10}\right)-\frac{76.64\times {10}^3}{2.303\times 8.314\times 318}$
= (0.6021+10) – 12.5876
= –1.9855
Therefore, k = Antilog (–1.9855)
= 1.034 ×$\times$ 10-2 s-1

---

30. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Ans. From Arrhenius equation, we obtain logk2k1=Ea2.303R(T2−T1T1T2)$\log \frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{R}}\left(\frac{{\mathrm{T}}_2-{\mathrm{T}}_1}{{\mathrm{T}}_1{\mathrm{T}}_2}\right)$
It is given that, k2 = 4k1
T1 = 293 K
T2 = 313 K
Therefore, log4k1k2=Ea2.303×8.314(313−293293×313)$\log \frac{4{\mathrm{k}}_1}{{\mathrm{k}}_2}=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times 8.314}\left(\frac{313-293}{293\times 313}\right)$
0.6021=20×E22.303×8.314×293×313$0.6021=\frac{20\times {\mathrm{E}}_2}{2.303\times 8.314\times 293\times 313}$
Ea=0.6021×2.303×8.314×293×31320${\mathrm{E}}_a=\frac{0.6021\times 2.303\times 8.314\times 293\times 313}{20}$
= 52863.33 J mol-1
= 52.86 kJ mol-1
Hence, the required energy of activation is 52.86 kJ mol-1.