Question

The Arrhenius equation for trans isomerisation of 2-butene and 2-butene nitrile are given as follows:
(a) For 2-butene $C{H}_{3}CH=CHC{H}_{3}K\left(S^{-1}\right)=10^{13.8}$
(b) For 2-butene nitrile $C{H}_{3}CH=CH.CN$ $K^{\prime }(S^{-1})=10^{11}ex{p}^{-214-5kJmo{l}^{-1}}/RT$.
The temperature at which $K=K^{\prime }$

• A. $913.87K$
• B. $533.43K$
• C. $1000.02K$
• D. $407.05K$

Answer 1

Answer: (A)

$\because K=K^{\prime }$
$10^{13.8}ex{p}^{-263.5kJmo{l}^{-1}RT}=10^{11}ex{p}^{-214.5mo{l}^{-1}RT}$
or $10^{13.8-11}=exp\frac{\left[\left(-214.5\right)-\left(-263-5\right)\right]kJmo{l}^{-1}}{8.314\times 10^{-3}kJmo{l}^{-1}{K}^{\circ }\times T}$
or $630.957=exp\frac{5.893\times 10^3}{T}$
$\therefore 2.303log630.957=\frac{5.893\times 10^3}{T}$
$\therefore T=\frac{5.893\times 10^3}{2.303\times 2.8}=913.87K$