Question

The plot of \(\log K\) versus \(\dfrac{1}{T}\) is a straight line. The intercept and slope of this line are respectively given by (Where \(K\) is the equilibrium constant.)

• A. \(\dfrac{\Delta S^\circ}{2.303R},\ -\dfrac{\Delta H^\circ}{2.303R}\)
• B. \(\dfrac{\Delta S^\circ}{R},\ -\dfrac{\Delta H^\circ}{R}\)
• C. \(-\dfrac{\Delta S^\circ}{2.303R},\ \dfrac{\Delta H^\circ}{2.303R}\)
• D. \(-\dfrac{\Delta H^\circ}{2.303R},\ \dfrac{\Delta S^\circ}{2.303R}\)

Hint:

For \(\log K\) vs \(1/T\) plots, intercept gives \(\Delta S^\circ\) and slope gives \(\Delta H^\circ\) — both divided by \(2.303R\).

Answer 1

The Correct Option is A

Concept:
The temperature dependence of the equilibrium constant is given by the van’t Hoff equation
: \[ \ln K = -\frac{\Delta H^\circ}{RT} + \frac{\Delta S^\circ}{R} \]
Step 1: Convert Natural Logarithm to Base-10 Logarithm
Using: \[ \ln K = 2.303 \log K \] \[ \Rightarrow \log K = -\frac{\Delta H^\circ}{2.303R}\left(\frac{1}{T}\right) + \frac{\Delta S^\circ}{2.303R} \]
Step 2: Compare with Straight Line Equation
General straight line form: \[ y = mx + c \] Here: \[ y = \log K, \quad x = \frac{1}{T} \] Thus, \[ \text{Slope } (m) = -\frac{\Delta H^\circ}{2.303R} \] \[ \text{Intercept } (c) = \frac{\Delta S^\circ}{2.303R} \]
Final Conclusion:
\[ \boxed{\text{Intercept} = \frac{\Delta S^\circ}{2.303R}, \quad \text{Slope} = -\frac{\Delta H^\circ}{2.303R}} \]