Question
The enthalpy of combustion of C, H _2, and CH _4 are -390, -285 and -890 kJ/mol. Find the enthalpy of formation of methane.
MHT CET
Chemistry
Thermodynamics (C)
2026
MHT CET 2026 (20 April Shift 2)
MHT CET Chemistry Question (2026) — Solution
Options
- A. -70 kJ
- B. -111 kJ
- C. -170 kJ
- D. -85 kJ
Answer
A. -70 kJ
Step-by-step solution
The required reaction for the formation of methane is: C(s) + 2 H _2 (g) CH _4 (g) The enthalpy of formation can be calculated using the enthalpies of combustion of the reactants and products: H_f( CH _4) = H_c( C ) + 2 H_c( H _2) - H_c( CH _4) Substituting the given values: H_f( CH _4) = -390 + 2(-285) - (-890) H_f( CH _4) = -390 - 570 + 890 H_f( CH _4) = -960 + 890 = -70 kJ/mol Answer: -70 kJ
Related: Chemistry — Thermodynamics (C) · All PYQ Banks