Question

For the given reaction, $C_{2}H4\left(g\right)+H_2\left(g\right)\leftrightharpoons C_2{H}_6\left(g\right)$, $\Delta H$ = (-)ve
It is carried out in a vessel. The equilibrium concentration of $C_2{H}_4$ can be decreased by :

• A. increasing the temperature
• B. increasing the pressure
• C. removing some $H_2$
• D. adding some $C_2{H}_6$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A increasing the temperature
B increasing the pressure
C removing some $H_2$
D adding some $C_2{H}_6$

$\begin{array}{c}{\text{C}}_2{H}_4\left(g\right)+H_2\left(g\right)\rightleftharpoons C_2{H}_6\left(g\right);\Delta H=-32.7kcal\\ \text{concentration of}{C}_2{H}_4\text{will increase by all the}\\ \text{factors favouring backward reaction. High}\\ \text{temperature, low pressure, decrease in}\\ \text{concentration of}{H}_2\text{and increase in concentration}\\ \text{of}{C}_2{H}_6\text{will favour backward reaction.}\\ \therefore \left(a\right),\left(b\right),\left(c\right)\text{and}\left(d\right)\text{all correct.}\end{array}$