Calculate G^0 for reaction, CH_ 4(g) + H_ 2(g) C_2H_ 6(g) (K_P = 2 x 10^ 17 , R = 8.314 J K^ -1 mol^ -1 )

MHT CET 2026 Chemistry question

Calculate for reaction, ( = 2 x 10, R = 8.314 J Kmol)

(A)-64.695 kJ mol
(B)-98.716 kJ molCorrect
(C)-44.08 kJ mol
(D)-58.78 kJ mol

Answer: (B)

Solution

The standard Gibbs energy change is related to the equilibrium constant by . Taking standard temperature K (not printed in the question): . So G^0$ = -2.303 × 8.314 × 298 × 17.301 ≈ -5706

See the full step-by-step solution

Create a free account to read the complete working, and practise questions like this in a timed test.

Create a free account

Question ID VK-101612