Chemical and Ionic Equilibrium - Result Question 35
35. For the gas phase reaction,
$ \mathrm{C}_2 \mathrm{H}_4+\mathrm{H}_2 \rightleftharpoons \mathrm{C}_2 \mathrm{H}_6 \quad(\Delta H=-32.7 \mathrm{~k cal}) $
carried out in a vessel, the equilibrium concentration of $\mathrm{C}_2 \mathrm{H}_4$ can be increased by
(1984, 1M)
(a) increasing the temperature
(b) decreasing the pressure
(c) removing some $\mathrm{H}_2$
(d) adding some $\mathrm{C}_2 \mathrm{H}_6$
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Answer:
Correct Answer: 35. (a, b, c, d)
Solution:
- $C _2 H _4+H _2 \rightleftharpoons C _2 H _6, \Delta H=-32.7 $ $kcal$
The above reaction is exothermic, increasing temperature will favour backward reaction, will increase the amount of $C _2 H _4$. Decreasing pressure will favour reaction in direction containing more molecules (reactant side in the present case). Therefore, decreasing pressure will increase amount of $C _2 H _4$.
Removing $H _2$, which is a reactant, will favour reaction in backward direction, more $C _2 H _4$ will be formed.
Adding $C _2 H _6$ will favour backward reaction and some of the $C _2 H _6$ will be dehydrogenated to $C _2 H _4$.
BETA
