Question

The plot of log K against $\frac{1}{T}$ is a straight line with positive slope (K being the equilibrium constant of a reaction), which of the following is then correct?

• A. The reaction is endothermic in nature
• B. The reaction will be exothermic in nature
• C. The reaction goes to farther extent on raising the temperature
• D. None of these

Answer 1

Answer: (B)

The reaction will be exothermic in nature

Since the slope of the straight line graph between $logK$ and $\frac{1}{T}$ is positive, $\Delta H_{reaction}$ would be negative, i.e., the forward reaction would be exothermic.