Question

If the change in entropy of the reaction is 0.07 kJ. K${}^{-1}$ mol${}^{-1}$ at 1 atm pressure. Calculate up to which temperature the reaction would not be spontaneous.

$[$For forward reaction $\Delta H=20\text{kJ/mol}]$.

• A. T < 285.7 K
• B. T > 250 K
• C. T < 340.2 K
• D. T > 200 K

Answer 1

The correct option is A T < 285.7 K

The relationship between the free energy change, enthalpy change and entropy change is $\Delta G=\Delta H-T\Delta S$.

$\Delta G=20-T\times 0.07$

When $\Delta G>0$, the process is non-spontaneous.

Thus, $20-T\times 0.07>0$

or $T<\frac{20}{0.07}$

or $T<285.7K$.

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