Kelvi

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)

medium

Thermodynamics

2017

chemistry

T > 425 K

All temperatures

T > 298 K

T < 425 K

Explanation

To determine the spontaneity of a reaction, we use the Gibbs free energy change $\Delta G. \\$ The formula for $\Delta G$ is: $\\ \Delta G = \Delta H - T \Delta S \\$ For a reaction to be spontaneous, $\Delta G$

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

Explanation

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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)