Question

A straight line having a slope of $-\Delta U^{0}/R$ is obtained in a plot between

• A. lnK$_p$ versus T
• B. lnK$_c$ versus T
• C. lnK$_p$ versus 1/T
• D. lnK$_c$ versus 1/T

Hint:

Remember: lnK vs T plots relate to $\Delta U^0$, while lnK vs 1/T plots relate to $\Delta H^0$.

Answer 1

The Correct Option is D

Step 1: Understanding relation.
The temperature dependence of K$_p$ is given by:
$\frac{d(\ln K_p)}{dT} = \frac{\Delta U^0}{RT^2}$. On integrating, $\ln K_p = -\frac{\Delta U^0}{R} \frac{1}{T} + \text{constant}$. But for internal energy change, the temperature-based form gives slope proportional to $-\Delta U^0 / R$ with lnK$_p$ vs T.
Step 2: Identifying the correct plot.
When slope = $-\Delta U^0 / R$, the plot must be lnK$_p$ versus T.
Step 3: Conclusion.
Hence, the required plot is lnK$_p$ vs T.