The plot of _ 10 ~K vs 1 ~T gives a straight line. The intercept and slope respectively are (where K is equilibrium constant).

Question

Quantrex Academy · Accepted Answer

Correct answer: A. Δ S^ o 2 · 303 R ,- Δ H^ o 2 · 303 R

The relationship between the standard Gibbs free energy change and the equilibrium constant is given by

\Delta G^o = -RT \ln K

.
We also know that

\Delta G^o = \Delta H^o - T\Delta S^o

.
Equating the two expressions:

-RT \ln K = \Delta H^o - T\Delta S^o

.
Dividing by

-RT

, we get:

\ln K = -\frac{\Delta H^o}{RT} + \frac{\Delta S^o}{R}

.
To convert natural log to base 10, use

\ln K = 2.303 \log_{10} K

:

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

.
Dividing by

2.303

:

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

.
Rearranging in the form of a straight line equation

y = mx + c

where

y = \log_{10} K

and

x = \frac{1}{T}

:

\log_{10} K = \left( -\frac{\Delta H^o}{2.303R} \right) \frac{1}{T} + \frac{\Delta S^o}{2.303R}

.
Comparing the terms:
Intercept (

c

) =

\frac{\Delta S^o}{2.303R}

Slope (

m

) =

-\frac{\Delta H^o}{2.303R}

.

The plot of _ 10 ~K vs 1 ~T gives a straight line. The intercept...