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MHT CET202611 April 2026Evening ShiftChemistryThermodynamics (C)Actual

The equilibrium constant for a reaction is 100 . What will be the value of standard Gibbs energy change at 298 K ? ( R = 8.314 J K ⁻¹ mol ⁻¹ )

Options

  1. A-11.411 KJ/mol
  2. B-5.744 KJ/mol
  3. C-570.584 KJ/mol
  4. D-57.058 KJ/mol

Correct answer

A. -11.411 KJ/mol

Step-by-step solution

The standard Gibbs free energy change is given by the equation: G^ = -RT K = -2.303 RT K Substituting the given values: R = 8.314 J K ⁻¹ mol ⁻¹ T = 298 K K = 100 G^ = -2.303 8.314 298 (100) G^ = -2.303 8.314 298 2 G^ = -11411.7 J mol ⁻¹ G^ -11.411 kJ mol ⁻¹ Answer: -11.411 KJ/mol

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