Question

The plot of \(\log_{10}K\) vs \(\frac{1}{T}\) gives a straight line. The intercept and slope respectively are (where \(K\) is equilibrium constant).

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

Whenever you see \(\log K\) vs \(1/T\), immediately recall the Van’t Hoff equation.

Answer 1

The Correct Option is C

Concept: From thermodynamics, the Van’t Hoff equation is: \[ \ln K = -\frac{\Delta H^\circ}{RT} + \frac{\Delta S^\circ}{R} \] Converting natural logarithm to base 10: \[ \log_{10}K = -\frac{\Delta H^\circ}{2.303R}\cdot\frac{1}{T} + \frac{\Delta S^\circ}{2.303R} \]
Step 1: Identify slope Comparing with equation of straight line \(y=mx+c\): \[ \text{slope } m = -\frac{\Delta H^\circ}{2.303R} \]
Step 2: Identify intercept \[ \text{intercept } c = \frac{\Delta S^\circ}{2.303R} \] Final Answer: \[ \boxed{\text{Intercept }=\frac{\Delta S^\circ}{2.303R},\quad \text{Slope }=-\frac{\Delta H^\circ}{2.303R}} \]