Question

Variation of equilibrium constant K for the reaction ; $2A\left(s\right)+B\left(g\right)\rightleftharpoons C\left(g\right)+2D\left(g\right)$ is plotted against absolute T in figure as- $lnKVs\left(1/T\right)$ :

• A. the forward reaction is exothermic
• B. the forward reaction is endothermic
• C. the slope of line is proportional to $\Delta H$
• D. adding 'A' favours forward reaction
• E. removing C favours forward reaction

Answer 1

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

The correct options are
A the forward reaction is exothermic
C the slope of line is proportional to $\Delta H$
D adding 'A' favours forward reaction
E removing C favours forward reaction

The relation between Equilibrium constant and Temperature is given by Van't Hoff equation,
$ln{K}_{eq}=-\frac{\Delta H^0}{RT}+\frac{\Delta S^0}{R}$.
For the $ln{K}_{eq}vs\frac{1}{T}$ plot, $-\frac{\Delta H^0}{R}$ is the slope.
The graph suggests positive slope, hence $-\frac{\Delta H^0}{R}$ is positive or $\Delta H^0$ is negative. Hence, the reaction is exothermic. Slope line is proportional to $\Delta H$ Also, by Le Chatelier's principle, removing C favours forward reaction.