Question

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

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

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, $T>\frac{\Delta H}{\Delta S}=\frac{35.5kJmo{l}^{-1}}{83.6J{K}^{-1}mo{l}^{-1}}$
$=\frac{35.5\times 1000Jmo{l}^{-1}}{83.6J{K}^{-1}mo{l}^{-1}}=425K$
Thus, the reaction is spontaneous at
$T>425K$