Question

The equilibrium constant at ${427}^{\circ }C$ for the reaction: is $K_P=8\times {10}^{-5}$. Calculate the value of $\Delta G^{\circ }$ for the reaction at ${427}^{\circ }$.Take ln2 =0.3 and R = 8.3

• A. $-33\text{kJ}$
• B. $-54\text{kJ}$
• C. $54.7\text{kJ}$
• D. $55\text{J}$

Answer 1

The correct option is C $54.7\text{kJ}$

By using the formade
$\Delta G^{\circ }=-RTln{K}_p$
Where,
$R=8.3J{k}^{-1}{\text{mol}}^{-1}$
T = Temperature = 427 + 273 = 700 K
$K_P=8\times {10}^{-5}$ (given)
Substituting the value, we get,
$\Delta G^{\circ }=-8.3\times 700\times ln(2^3\times {10}^{-5})$
$\Delta G^{\circ }=-8.3\times 700\times \left(ln\right(2^3)+ln{10}^{-5})$
$=-8.3\times 700\times \left(ln\right(2^3)+ln{10}^{-5})$
$=-8.3\times 700\times (2.07-11.5)$
$=5.47\times {10}^4{\text{Jmol}}^1$

$=5.47{\text{kJ mol}}^{-1}$
Hence, option (C) is correct.