Question

For a particular reaction at 111.6 °C, ΔG = 131.35 kJ/mol, and ΔS = 201.60 J/(mol·K)....

For a particular reaction at 111.6 °C, ΔG = 131.35 kJ/mol, and ΔS = 201.60 J/(mol·K).

Calculate ΔG for this reaction at -25.3 °C.

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Answer #1

use the formula

ΔG = ΔH - TΔS calculate the first ΔH

ΔG = 131.35 kJ/mol convert in to joules = 131350 J/mol

ΔS = 201.60 J/(mol·K)

T = 273 + 111.6ºC = 384.6 K

131350 J/mol = ΔH - 384.6 K (201.60 J/(mol·K))

ΔH = 131350  J/mol + 77535.36 J/mol

ΔH = 208885.36 J/mol

now use this ΔH value and calculate the ΔG value

ΔG = ΔH - TΔS

ΔH = 208885.36 J/mol

ΔS = 201.60 J/(mol·K)

T = 273 - 25.3ºC = 247.7K

ΔG = 208885.36 J/mol - 247.7K (201.60 J/(mol·K))

ΔG = 208885.36 J/mol - 49936.32 J/mol

ΔG = 158949.04 J/mol convert in to kJ/mol by dividing with 1000

ΔG = 158.949 kJ/mol

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