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., $\text{T}>425\text{K}$