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

The given problem deals with the thermodynamic relationship related to the equilibrium constant \( K \) and its dependence on temperature, which can be expressed using the Van't Hoff equation. The equation is:

\[\log_{10} K = \frac{-\Delta H^\circ}{2.303RT} + \frac{\Delta S^\circ}{2.303R}\]

This equation represents a linear relationship between \(\log_{10} K\) and \(\frac{1}{T}\), where:

• \(-\frac{\Delta H^\circ}{2.303R}\) is the slope of the line.
• \(\frac{\Delta S^\circ}{2.303R}\) is the intercept of the line.

Let's analyze why the correct answer is \(\frac{\Delta S^\circ}{2.303R},\ -\frac{\Delta H^\circ}{2.303R}\):

1. The Van't Hoff equation is derived from the Gibbs free energy relationship and the definition of equilibrium constant: \(\Delta G^\circ = -RT \ln K\).
2. This is further expressed in terms of enthalpy and entropy changes using: \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\).
3. Rearranging and substituting into the logarithmic form gives the linear equation depicting the plot between \(\log_{10} K\) vs \(\frac{1}{T}\), which results in the slope and intercept derived from this expression.

Therefore, the intercept of the plot is given by \(\frac{\Delta S^\circ}{2.303R}\), and the slope is \(-\frac{\Delta H^\circ}{2.303R}\). Hence, the correct option is:

\(\frac{\Delta S^\circ}{2.303R},\ -\frac{\Delta H^\circ}{2.303R}\)

By understanding this thermodynamic relationship, one can better grasp how changes in enthalpy and entropy affect the equilibrium constant \( K \) with temperature.