Question

A reversible reaction proceeds in the direction that decreases the free energy of the system. There is an important relationship between the equilibrium constant and the standard free energy of the system given by
$\text{log K}=\frac{-\Delta G^0}{2.303}\text{RT}$
This equation is applicable to all reversible processes and allows us to estimate the equilibrium constant from the standard free energy and the enthalpy change of the reaction.
Take $R=\frac{25}{3}\text{J/molK,}\frac{1}{12}\text{L atm/molk,}\ln 2=0.7,\ln 3=1.1,\ln 5=1.6,\text{log}2=0.3,\text{log 3}=0.5,log5=0.7$

The equilibrium constant for the dissociation of $N_2{O}_4$ into $N{O}_2$ is 0.125 at ${27}^{\circ }$C. What is the standard free energy change for this reaction?

• A. $-5.25\text{kJ/mole}$
• B. $+5.25\text{kJ/mole}$
• C. $+22.50\text{kJ/mole}$
• D. $-22.50\text{kJ/mole}$

Answer 1

The correct option is B $+5.25\text{kJ/mole}$

$\Delta G=-R\times T\times lnK$
$\Delta G=-\frac{25}{3}\times 300\times ln0.125$
$\Delta G=-\frac{25}{3}\times 300\times ln\frac{1}{8}$
$\Delta G=-\frac{25}{3}\times 300\times (ln1-ln8)$
$\Delta G=-\frac{25}{3}\times 300\times (0-3ln2)$
$\Delta G=5.25kJmo{l}^{-1}$