Chemical Kinetics - Result Question 3
3. For the reaction of $H _2$ with $I _2$, the rate constant is $2.5 \times 10^{-4} dm^{3} mol^{-1} s^{-1}$ at $327^{\circ} C$ and $1.0 dm^{3} mol^{-1} s^{-1}$ at $527^{\circ} C$. The activation energy for the reaction, in $kJ mol^{-1}$ is $\left(R=8.314 JK^{-1} mol^{-1}\right)$
(2019 Main, 10 April II)
(a) 59
(b) 72
(c) 150
(d) 166
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Answer:
Correct Answer: 3. (d)
Solution:
The Arrhenius equation for rate constants at two different temperatures is
$ \log \dfrac{k _2}{k _1}=\dfrac{E _a}{2.303 R}\left[\dfrac{T _2-T _1}{T _1 T _2}\right] \quad\left[\text { where, } T _2>T _1\right] $
where, $k _1$ and $k _2$ are rate constants at temperatures $T _1$ and $T _2$, respectively.
$R=$ Gas constant, $E _a=$ Activation energy
For the reaction, $H _2+I _2 \longrightarrow 2 HI$
Given $k _1=2.5 \times 10^{-4} dm^{3} mol^{-1} s^{-1}$
$T _1=(273+327) K=600 K$
$k _2=1 dm^{3} mol^{-1} s^{-1}$ at $T _2=(273+527) K=800 K$
Now, $\log \dfrac{k _2}{k _1}=\dfrac{E _a}{2.303 R}\left(\dfrac{T _2-T _1}{T _1 T _2}\right)$
$\Rightarrow \log \dfrac{1}{2.5 \times 10^{-4}}=\dfrac{E _a}{2.303 \times 8.314 \times 10^{-3}}\left(\dfrac{800-600}{600 \times 800}\right)$
$\Rightarrow \log \dfrac{\left(10 \times 10^{3}\right)}{2.5}=\dfrac{E _a}{0.019} \times \dfrac{200}{48 \times 10^{4}}$
$\Rightarrow \log 4+3 \log 10 \simeq E _a \times 0.022$
$\Rightarrow \quad E _a=\dfrac{2 \times \log 2+3}{0.022}$
$ =\dfrac{3.6}{0.022} \simeq 163.6 kJ \hspace{1mm} mol^{-1} $
BETA
