For a particular reaction at 129.9 °C, ΔG = -603.84 kJ/mol, and ΔS = 313.84 J/(mol·K).Calculate...

For a particular reaction at 129.9 °C, ΔG = -603.84 kJ/mol, and ΔS = 313.84 J/(mol·K).Calculate ΔG for this reaction at -3.4 °C.

Solutions

Expert Solution

T1 = 129.9^oC = 129.9 + 273.15 = 403.05K

T₂ = -3.4^oC = -3.4 + 273.15 = 269.75K

ΔG = -603.84 kJ/mol, and ΔS = 313.84 J/(mol·K) at T₁ = 403.05K

calculating H for the reaction

G = H - TS

So, H = G + TS = (-603.84 kJ/mol) + (403.05K)(313.84 J/(mol·K) )

= (-603.84 kJ/mol) + (126.49 kJ/mol)

H = - 477.35 kJ/mol

To calculate G at -3.4^oC

T = -3.4^oC = -3.4 + 273.15 = 269.75K

ΔS = 313.84 J/(mol·K) and H = - 477.35 kJ/mol

G = H - TS

G = (- 477.35 kJ/mol) - (269.75K)(313.84 J/(mol·K))

= (- 477.35 kJ/mol) - (84.66kJ/mol)

G = - 562.01 kJ/mol at -3.4^oC

queen_honey_blossom answered 1 year ago

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