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 $\Delta H=-v e$