Question

The enthalpy change for a reaction at equilibrium is $-20.5$ kJ mol${}^{-1}$. Then the entropy change for this equilibrium at 410 K is:

• A. + 50 JK${}^{-1}$ mol${}^{-1}$
• B. + 55 JK${}^{-1}$ mol${}^{-1}$
• C. + 75 JK${}^{-1}$ mol${}^{-1}$
• D. - 50 JK${}^{-1}$ mol${}^{-1}$
• E. - 55 JK${}^{-1}$ mol${}^{-1}$

Answer 1

Answer: (D)

- 50 JK${}^{-1}$ mol${}^{-1}$

Given, the enthalpy change

$=-20.5$ kJ / mol

$=-20.5\times {10}^3$ J/mol

Temperature $\left(T\right)=410$ K

Entropy change

$\Delta S=\frac{\Delta H}{T}=\frac{(-20.6\times {10}^3)}{410}$

$=-50$ JK${}^{-1}$ mol${}^{-1}$

Hence, the correct option is $\text{D}$