For a particular reaction ΔH=−38.3 kJ and ΔS∘=−113J×K−1 This reaction is :

ΔH−TΔS ΔHΔS⇒T<38.3×103113=66∘C

For a particular reaction ΔH=−38.3 kJ and ΔS∘=−113J×K−1 This reaction is :

ΔH−TΔS ΔHΔS⇒T<38.3×103113=66∘C

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For a particular reaction $Δ H = - 38.3 kJ$ and $Δ S^{\circ} = - 113 J \times K^{- 1}$ This reaction is :

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spontaneous at all temperatures.

non-spontaneous at all temperatures

spontaneous at temperatures below 66° C.

spontaneous at temperatures above 66° C

answer is C.

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Detailed Solution

$\begin{array}{l} Δ H - T Δ S < 0 \\ \Rightarrow T > \frac{Δ H}{Δ S} \\ \Rightarrow T < \frac{38.3 \times 10^{3}}{113} = 66^{\circ} C \end{array}$

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For a particular reaction ΔH=−38.3 kJ and ΔS∘=−113J×K−1 This reaction is :

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For a particular reaction $Δ H = - 38.3 kJ$ and $Δ S^{\circ} = - 113 J \times K^{- 1}$ This reaction is :