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IB Chemistry Test Prep

Category - IB Chemistry

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The following equation is under thermodynamic control. Assuming a temperature of 273K and using the provided information, find the Keq:

A + B --> C

DGf° (A) = -200 kJ/mol
DGf° (B) = -100 kJ/mol
DGf° (C) = -50 kJ/mol
  1. Keq = 1.46 * 10-14
  2. Keq = 1.46 * 10-38
  3. Keq = 1.46 * 10-48
  4. Keq = 0.83 * 10-27
  5. Keq = 0.83 * 10-32
Explanation
Answer: C. Solution is as follows:

Under thermodynamic control, the following equation applies:

ln (Keq) = -DGrxn°/(R*T)
R=ideal gas constant = 8.314 J/mol K

DGrxn°=DGproducts° - DGreactants°
DGrxn° = [-50 kJ/mol * 1mol] - [-100kJ/mol * 1mol - 200kJ/mol * 1 mol] = 250 kJ = 250,000J

ln (Keq) = -(250,000J)/((8.314J/mol K)*(273K))
solving for Keq = 1.46 * 10-48
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