Question
Calculate H for the following reaction at 300\ K 2 C _ (s) + 3 H _ 2(g) C _2 H _ 6(g) if U for the reaction is -80\ kJ ( R = 8.314\ JK ^ -1 mol ^ -1 )
MHT CET
Chemistry
Thermodynamics (C)
2026
MHT CET 2026 (18 April Shift 1)
MHT CET Chemistry Question (2026) — Solution
Options
- A. -85.00\ kJ
- B. -43.00\ kJ
- C. -128.00\ kJ
- D. -170.00\ kJ
Answer
A. -85.00\ kJ
Step-by-step solution
The given reaction is 2 C _ (s) + 3 H _ 2(g) C _2 H _ 6(g) Change in number of gaseous moles, n_g = n_ p(g) - n_ r(g) = 1 - 3 = -2 Given U = -80\ kJ T = 300\ K R = 8.314\ J K ^ -1 mol ^ -1 = 8.314 10^ -3 \ kJ K ^ -1 mol ^ -1 Using the relation H = U + n_g RT H = -80 + (-2) 8.314 10^ -3 300 H = -80 - 4.9884 H = -84.9884\ kJ -85.00\ kJ Answer: -85.00\ kJ
Related: Chemistry — Thermodynamics (C) · All PYQ Banks