Question
Calculate G^ for reaction, CH _ 4(g) + H _ 2(g) C _2 H _ 6(g) (K_p = 2 10^ 17 , R = 8.314\ JK ^ -1 mol ^ -1 )
MHT CET
Chemistry
Chemical Equilibrium
2026
MHT CET 2026 (20 April Shift 1)
MHT CET Chemistry Question (2026) — Solution
Options
- A. -64.695 kJ mol^ -1
- B. -98.716 kJ mol^ -1
- C. -44.08 kJ mol^ -1
- D. -58.78 kJ mol^ -1
Answer
B. -98.716 kJ mol^ -1
Step-by-step solution
The standard Gibbs free energy change is related to the equilibrium constant by the equation: G^ = -RT K_p = -2.303 RT K_p Assuming standard temperature T = 298 K and substituting the given values: G^ = -2.303 8.314 298 (2 10^ 17 ) G^ = -5705.84 ( 2 + 17 10) G^ = -5705.84 (0.3010 + 17) G^ = -5705.84 17.3010 G^ = -98716 J mol ^ -1 G^ = -98.716 kJ mol ^ -1 Answer: -98.716 kJ mol^ -1
Related: Chemistry — Chemical Equilibrium · All PYQ Banks