At 500 K for an isobaric process, ΔSsystem=−10kJmol K and ΔSsurr=12kJmol K Therefore,ΔG for the entire process is :

ΔStotal=−10+12=2 kJ/mol−KΔG=−TΔStotal=−500×2=−1000 kJ/mol

At 500 K for an isobaric process, ΔSsystem=−10kJmol K and ΔSsurr=12kJmol K Therefore,ΔG for the entire process is :

ΔStotal=−10+12=2 kJ/mol−KΔG=−TΔStotal=−500×2=−1000 kJ/mol

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At 500 K for an isobaric process, $Δ S_{s y s t e m} = - 10 \frac{k J}{m o l K} and Δ S_{s u r r} = 12 \frac{k J}{m o l K}$ Therefore, $Δ G$ for the entire process is :

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-500 kJ/mol

-1000 kJ/mol

-600 kJ/mol

-1100 kJ/mol

answer is B.

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$Δ S_{t o t a l} = - 10 + 12 = 2 k J / m o l - K$

$Δ G = - T Δ S_{t o t a l} = - 500 \times 2 = - 1000 k J / m o l$

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At 500 K for an isobaric process, ΔSsystem=−10kJmol K and ΔSsurr=12kJmol K Therefore,ΔG for the entire process is :

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At 500 K for an isobaric process, $Δ S_{s y s t e m} = - 10 \frac{k J}{m o l K} and Δ S_{s u r r} = 12 \frac{k J}{m o l K}$ Therefore, $Δ G$ for the entire process is :