Question

Calculate free energy change for the following reaction at $298K$:
$2NO\left(g\right)+B{r}_2\left(l\right)\to 2NOBr\left(g\right)$
Given the partial pressure of $NO$ is $0.1\text{atm}$ and the partial pressure of $NOBr$ is $2.0atm$ and $\Delta G_{fNOBr}^0=82.4{\text{kJ mol}}^{-1},\Delta G_{NO}^0=86.55{\text{kJ mol}}^{-1}$

• A. $6.5\text{kJ}$
• B. $-8.3\text{kJ}$
• C. $-1.4\text{kJ}$
• D. $+8.3\text{kJ}$

Answer 1

The correct option is A $6.5\text{kJ}$

The reaction given is,
$2NO\left(g\right)+B{r}_2\left(l\right)\to 2NOBr\left(g\right)$
$\therefore \Delta G^{\circ }=2\left[\Delta G_f^{0}NOBr\right]-\left[2\right[\Delta G_f^{0}NO]+\Delta G_f^0[B{r}_2\left]\right]=2\times 82.4-2\times 86.55+1\times 0=-8.3\text{kJ}=-8.3\times {10}^3\text{J}$
Now, $2NO\left(g\right)+B{r}_2\left(l\right)\to 2NOBr{Q}_p=\frac{(P_{NOBr}{)}^2}{(P_{NO}{)}^2}=\frac{2^2}{(0.1{)}^2}=4\times {10}^2$
We know,
$\Delta G=\Delta G^0+RTln{Q}_p=-8.3\times {10}^3+(8.314\times 298\times 2.303)log(4\times {10}^2)=-8.3\times {10}^3+1.48\times {10}^4\text{J}=+6.5\times {10}^3=6.5\text{kJ}$