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MHT CET Chemistry Thermodynamics (C) 2026 MHT CET 2026 (19 April Shift 2)

MHT CET Chemistry Question (2026) — Solution

Question

Calculate standard Gibbs energy for any gaseous reaction if values of K_p and R respectively are 3.5 10^ 17 ; 8.314\, JK ^ -1 mol ^ -1

Options

  1. A. -100.1\, kJ mol ^ -1
  2. B. -79.5\, kJ mol ^ -1
  3. C. -71.4\, kJ mol ^ -1
  4. D. -89.5\, kJ mol ^ -1

Answer

A. -100.1\, kJ mol ^ -1

Step-by-step solution

The standard Gibbs free energy change is given by the relation: G^ = -RT K_p = -2.303 RT K_p Assuming standard temperature T = 298\, K : G^ = -2.303 8.314 298 (3.5 10^ 17 ) G^ = -5705.8 ( 3.5 + 17) G^ = -5705.8 (0.544 + 17) G^ = -5705.8 17.544 G^ = -100102\, J mol ^ -1 -100.1\, kJ mol ^ -1 Answer: -100.1\, kJ mol ^ -1

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