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

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