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Problems · Problem 4.17

Q.For a certain reaction ΔH0\Delta H^0 is -224 kJ and ΔS0\Delta S^0 is -153 J K−1^{-1}. At what temperature the change over from spontaneous to nonspontaneous will occur?

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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✓ Free question

T=ΔH0/ΔS0=(−224)/(−0.153)=1464T = \Delta H^0/\Delta S^0 = (-224)/(-0.153) = 1464 K; spontaneous below, nonspontaneous above.

Step 1. At the changeover temperature ΔG=ΔH0−TΔS0=0\Delta G = \Delta H^0 - T\Delta S^0 = 0, so T=ΔH0ΔS0T = \dfrac{\Delta H^0}{\Delta S^0}.

Step 2. Match the units: ΔS0=−153\Delta S^0 = -153 J K−1^{-1} =−0.153= -0.153 kJ K−1^{-1}.

Step 3. T=−224 kJ−0.153 kJ K−1=+1464T = \dfrac{-224\ \mathrm{kJ}}{-0.153\ \mathrm{kJ\ K^{-1}}} = +1464 K.

Step 4. With ΔH0<0\Delta H^0 < 0 and ΔS0<0\Delta S^0 < 0, the −TΔS0-T\Delta S^0 penalty grows with temperature, so the reaction is spontaneous below 1464 K and nonspontaneous above it.

✓Final answer

T=1464T = 1464 K; the reaction is spontaneous below 1464 K -- digit-for-digit the textbook's printed final.

Note

The textbook's printed quotient keeps the unit label "J K−1^{-1}" against the converted value (it prints T=−224 kJ/−0.153T = -224\ \mathrm{kJ} / -0.153 J K−1^{-1}) -- the value −0.153-0.153 is in kJ K−1^{-1}, as used above.

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