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Question: At 500 K for an isobaric process. DS<sub>system</sub> = – 10 \(\frac{kJ}{molK}\) and DS<sub>surr</s...

At 500 K for an isobaric process.

DSsystem = – 10 kJmolK\frac{kJ}{molK} and DSsurr = 12 kJmolK\frac{kJ}{molK}

Therefore, DG for the entire process is-

A

–500 kJ/mol

B

–1000 kJ/mol

C

–600 kJ/mol

D

–1100 kJ/mol

Answer

–1000 kJ/mol

Explanation

Solution

DStotal = DSsystemΔHT\frac{\Delta H}{T} = 2 kJ/mol.

or ΔHTΔST\frac{\Delta H–T\Delta S}{T} = – 2 kJ/mol.

or DH – TDS = – 2T kJ/mol. = – 2 × 500 kJ/mol = – 1000

kJmol\frac{kJ}{mol}