An isothermal and isobaric process is accompanied by changes in enthalpy and entropy as ΔH = 52 kJ mol^-1 and ΔS = 165 JK^-1 mol^-1, respectively. Predict whether the process be spontaneous at 400K.
An isothermal and isobaric process is accompanied by changes in enthalpy and entropy as 52 kJ mol-1 and 165 JK-1 mol-1 , respectively. Predict whether the process be spontaneous at 400K.
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To determine if the process is spontaneous at a given temperature, we can use the Gibbs free energy change (\(\Delta G\)), which is related to the changes in enthalpy (\(\Delta H\)) and entropy (\(\Delta S\)) through the equation:
\[
\Delta G = \Delta H – T\Delta S
\]
Where:
– \(\Delta H\) is the change in enthalpy,
– \(T\) is the temperature in Kelvin,
– \(\Delta S\) is the change in entropy.
Given:
– \(\Delta H = 52 \text{ kJ mol}^{-1} = 52000 \text{ J mol}^{-1}\) (since 1 kJ = 1000 J)
– \(\Delta S = 165 \text{ JK}^{-1} \text{ mol}^{-1}\)
– \(T = 400 \text{ K}\)
Plugging these values into the equation for \(\Delta G\):
\[
\Delta G = 52000 \text{ J mol}^{-1} – 400 \text{ K} \times 165 \text{ JK}^{-1} \text{ mol}^{-1}
\]
We can calculate \(\Delta G\) as follows:
\[
\Delta G = 52000 \text{ J mol}^{-1} – 66000 \text{ J mol}^{-1}
\]
\[
\Delta G = -14000 \text{ J mol}^{-1}
\]
Since \(\Delta G\) is negative, the process is spontaneous at 400 K.