Question

For a given reaction, $\text{Δ}\text{H}=35.5 \, \text{k}\text{J} \, \text{m}\text{o}\text{l}^{- 1}$ and $\text{Δ}\text{S}=83.6 \, \text{J}\text{K}^{- 1}\text{m}\text{o}\text{l}^{- 1}.$ The reaction is spontaneous at : (Assume that $\text{Δ}\text{H}$ and $\text{Δ}\text{S}$ do not vary with temperature)

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

Answer 1

Answer: (B)

$\because \, \, \, \text{Δ}\text{G}=\text{Δ}\text{H}-\text{T}\text{Δ}\text{S}$
For a reaction to be spontaneous, $\text{Δ}\text{G}= \, -\text{v}\text{e}$
i.e., $\text{Δ}\text{H} < \text{T}\text{Δ}\text{S}$
$\therefore \, \, \text{T}>\frac{\text{Δ} \text{H}}{\text{Δ} \text{S}}=\frac{35.5 \times 10^{3} \, \text{J}}{83.6 \, \text{J} \text{K}^{- 1}}$
i.e., $\text{T}>425 \, \text{K}$