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Given the information A+BC⟶2D⟶DΔH∘ΔH∘=706.3 kJΔS∘=320.0 J/K=478.0 kJΔS∘=−187.0 J/K A+B⟶2DΔ H∘=706.3 kJΔ S∘=320.0 J/KC⟶DΔ H∘=478.0 kJΔ S∘=−187.0 J/K calculate ΔG∘Δ G∘ at 298 K for the reaction

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ajeigbeibraheem

The Gibbs free energy ΔG for the equation is -456.512 kJ

The Gibbs free energy ΔG is determined by using the parameters;

ΔG = ΔH - TΔS

where;

  • ΔH = change in enthalpy of the reaction
  • ΔS = change in entropy of the reaction

From the given reaction, we have:

  • A + B →  2D       ΔH  = 706.3 kJ    ΔS = 320 J/K    ------ (1)
  •       C  →  D       ΔH  = 478 kJ        ΔS  = -187 J/K  ----- (2)

From equation (2), Let multiply equation (2) by 2 and reverse the equation, so we get:

  •  2D  → 2C         ΔH = (2 × -478) kJ   ΔS  = (2× 187) J/K  ----- (3)    

       

By adding both equations (1) and (3), we have:

  • A + B   →  2D       ΔH  = 706.3 kJ         ΔS = 320 J/K     ------ (1)
  •      2D  → 2C       ΔH = - 956  kJ        ΔS  =  374 J/K     ----- (3)          
  • A + B   → 2C        ΔH = -249  kJ           ΔS = 694 J/K = 0.694 kJ/K

Recall that:

ΔG = ΔH - TΔS

where;

  • T = temperature at 298 k

∴

ΔG = -249.7 kJ - (298k × 0.694 kJ/K)

ΔG = -249.7 kJ - 206.812 kJ

ΔG = -456.512 kJ

Therefore, we can conclude that the Gibbs free energy ΔG for the equation is -456.512 kJ

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