Question

For a given reaction, $\Delta H=35.5\, kJ\, mol ^{-1}$ and $\Delta S=83.6\, JK ^{-1} mol ^{-1}$. The reaction is spontaneous at (Assume that $\Delta H$ and $\Delta S$ do not vary with temperature)

• A. $T< 425\, K$
• B. $T >425\, K$
• C. all temperatures
• D. $T >298\, K$

Answer 1

Answer: (B)

According to Gibbs-Helmholtz equation,
Gibbs energy $(\Delta G)=\Delta H-T \Delta S$
where, $\Delta H=$ Enthalpy change
$\Delta S=$ Entropy change
$T=$ Temperature
For a reaction to be spontaneous,
$\Delta G < 0 \text {. }$
$\therefore$ Gibbs-Helmholtz equation becomes,
$\Delta G=\Delta H-T \Delta S< 0$
or, $\Delta H< T \Delta S$
or, $T >\frac{\Delta H}{\Delta S}=\frac{35.5\, kJ\, mol ^{-1}}{83.6\, JK ^{-1} mol ^{-1}}$
$=\frac{35.5 \times 1000\, J\, mol ^{-1}}{83.6\, J\, K ^{-1} mol ^{-1}}=425\, K$
Thus, the reaction is spontaneous at
$T > 425\, K$