The standard enthalpies of formation of CH₄ (g), CO₂ (g) and H₂ O(l) are -74.8 kJ mol ⁻¹ , -393.5 kJ mol ⁻¹ and -285.8 kJ mol ⁻¹ respectively. Then the enthalpy change for the given reaction in kJ mol ⁻¹ will be: 2 CH ₄(g) + 4 O ₂(g) 2 CO ₂(g) + 4 H ₂ O (l)

Step-by-step solution

The enthalpy of reaction is calculated using the standard enthalpies of formation of the products and reactants. _ r H^ = _ f H^ ( products ) - _ f H^ ( reactants ) For the given reaction: _ r H^ = [2 _ f H^ ( CO ₂) + 4 _ f H^ ( H ₂ O )] - [2 _ f H^ ( CH ₄) + 4 _ f H^ ( O ₂)] Substituting the given values: _ r H^ = [2(-393.5) + 4(-285.8)] - [2(-74.8) + 4(0)] _ r H^ = [-787.0 - 1143.2] - [-149.6] _ r H^ = -1930.2 + 149.6 = -1780.6 kJ mol ⁻¹ Answer: -1780.6