Question
Calculate the standard enthalpy change for the following reaction, 2 C _2 H _6 (g) + 7 O _2 (g) 4 CO _2 (g) + 6 H _2 O (l) Given, _f H^ ( C _2 H _6) = -85\ kJ mol ^ -1 _f H^ ( CO _2) = -390\ kJ mol ^ -1 _f H^ ( H _2 O ) = -285\ kJ mol ^ -1
MHT CET
Chemistry
Thermodynamics (C)
2026
MHT CET 2026 (18 April Shift 2)
MHT CET Chemistry Question (2026) — Solution
Options
- A. -2900\ kJ
- B. -3100\ kJ
- C. -3000\ kJ
- D. -3200\ kJ
Answer
B. -3100\ kJ
Step-by-step solution
The standard enthalpy of reaction is calculated using the formula: _r H^ = _f H^ ( products ) - _f H^ ( reactants ) For the given reaction: _r H^ = [4 _f H^ ( CO _2) + 6 _f H^ ( H _2 O )] - [2 _f H^ ( C _2 H _6) + 7 _f H^ ( O _2)] Substituting the given values: _r H^ = [4(-390) + 6(-285)] - [2(-85) + 7(0)] _r H^ = [-1560 - 1710] - [-170] _r H^ = -3270 + 170 = -3100\ kJ Answer: -3100\ kJ
Related: Chemistry — Thermodynamics (C) · All PYQ Banks