Question

The equilibrium constant for a reversible chemical reaction varies with $T$ as:

From this, it may be deduced that:

• A. the value of $K_p$ increases with increase in temperature
• B. the forward reaction gives out heat
• C. there are more moelcules on the right hand side of the chemical equation than on the left
• D. the reaction proceeds ten times faster at 450 K than 400 K

Answer 1

Answer: (B)

Variation of equilibrium constant with temperature can be express as:
$2.303\log \frac{K_2}{K_1}=\frac{\Delta H}{R}\left[\frac{T_2-T_1}{T_1\cdot T_2}\right]$
when $K_1>K_2$ and $T_1>T_2$
then $\Delta H=-ve$