Question

The standard free energy change $\left(\Delta G^{\circ }\right)$ for 50% dissociation of $N_2{O}_4$ into$N{O}_2$ at ${27}^{\circ }C$ and 1 atm pressure is $-xJmo{l}^{-1}$. The value of $x$ is (Nearest Integer)
$[\text{Given}:R=8.31J{K}^{–1}mo{l}^{–1},log(1.33)=0.1239,ln(10)=2.3]$

Answer 1

$N_2{O}_4\rightleftharpoons 2N{O}_2$
$t=010$
$t=t_{eq}1-0.52\times 0.5$

$P_{N_2{O}_4}=0.33atm$
$P_{N{O}_2}=0.66atm$

$K_p=\frac{(P_{N{O}_2}{)}^2}{P_{N_2{O}_4}}=\frac{(0.66{)}^2}{0.33}=1.33$

$\Delta G^{\circ }=-RTln{K}_p$
$=-8.31\times 300\times 2.3\times log1.33$
$\approx -710Jmo{l}^{-1}$