Question

At 5$7°\text{C}$, gaseous dinitrogen tetraoxide is $50\%$ dissociated. Calculate the standard free energy change per mole of $N_2{O}_4\left(g\right)$ at this temperature and at $1atm$. ($R=8.3{\text{JK}}^{–1}{\text{mol}}^{–1},\ln 10=2.3,\log 2=0.3,\log 3=0.48$)

$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$

• A. $-756{\text{J mol}}^{-1}$
• B. $-856{\text{J mol}}^{-1}$
• C. $-656{\text{J mol}}^{-1}$
• D. None of these

Answer 1

The correct option is A $-756{\text{J mol}}^{-1}$

$\begin{array}{cccc}& N_2{O}_4\left(g\right)& \rightleftharpoons & 2N{O}_2\left(g\right)\\ t=0& 1& & 0\\ t=eq.& 1-0.5& & 2\times 0.5\end{array}$
$K_P=\frac{P_{N{O}_2}^2}{P_{N_2{O}_4}}$
$P_{N_2{O}_4}=\frac{0.5}{1.5}\text{atm}{P}_{N{O}_2}=\frac{1}{1.5}\text{atm}$
$K_p=\frac{{\left(\frac{1}{1.5}\right)}^2}{\left(\frac{0.5}{1.5}\right)}=\frac{4}{3}$
$\Delta G^{\circ }=-2.3\times 8.3\times 330\times \log \left(\frac{4}{3}\right)=-756{\text{J mol}}^{-1}$