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NCERT Class XII Chemistry
Chapter - Chemical Kinetics
Questions with Solutions

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Question : 29 of 39
Marks: +1, -0
The time required for 10% completion of the 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 × 1010^{10} s−1^{-1}, calculate k at 318 K and Ea_{a}.
Solution:  
t=2.303k1log⁡[R]090100[R]0,t = \frac{2.303}{k_{1}} \log \frac{[R]_{0}}{\frac{90}{100}[R]_{0}},
t=2.303k2log⁡[R]075100[R]0t = \frac{2.303}{k_{2}} \log \frac{[R]_{0}}{\frac{75}{100}[R]_{0}}
t=2.303k1log⁡109,t = \frac{2.303}{k_{1}} \log \frac{10}{9},
t=2.303k2log⁡43t = \frac{2.303}{k_{2}} \log \frac{4}{3}
  =2.303k1log⁡109,\; = \frac{2.303}{k_{1}} \log \frac{10}{9},
  =2.303k2log⁡43\; = \frac{2.303}{k_{2}} \log \frac{4}{3}
⇒k2k1=log⁡43log⁡109\Rightarrow \frac{k_{2}}{k_{1}} = \frac{\log \frac{4}{3}}{\log \frac{10}{9}}
=log⁡1.333log⁡1.111= \frac{\log 1.333}{\log 1.111}
=0.12490.0457=2.733= \frac{0.1249}{0.0457} = 2.733
log⁡k2k1=Ea2.303R(T2−T1T1T2)\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} \left( \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right)
⇒log⁡2.733=Ea2.303×8.314(308−298298×308)\Rightarrow \log 2.733 = \frac{E_{a}}{2.303 \times 8.314} \left( \frac{308 - 298}{298 \times 308} \right)
Ea=2.303×8.314×308×29810×0.4367E_{a} = \frac{2.303 \times 8.314 \times 308 \times 298}{10} \times 0.4367
    =19.147×308×29810×0.4367\;\; = \frac{19.147 \times 308 \times 298}{10} \times 0.4367
      =76.75 kJ mol−1\;\;\; = 76.75 \ \text{kJ} \ \text{mol}^{-1}
ln⁡k=ln⁡A−EaRT\ln k = \ln A - \frac{E_{a}}{R T}
log⁡k=log⁡A−Ea2.303RT\log k = \log A - \frac{E_{a}}{2.303 R T}
=log⁡(4×1010)−76.75×10002.303×8.314×318= \log (4 \times 10^{10}) - \frac{76.75 \times 1000}{2.303 \times 8.314 \times 318}
=10.6021−767506088.746= 10.6021 - \frac{76750}{6088.746}
=10.6021−12.6051=−2.003= 10.6021 - 12.6051 = -2.003
k=k = Antilog (−2.003)=9.93×10−3(-2.003) = 9.93 \times 10^{-3}
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