Given (A) 2 CO g + O 2 g → 2 CO 2 g ΔH 1 o = - x kJ mol - 1 (B) C graphite + O 2 g → CO 2 g ΔH 2 o = - y kJ mol - 1 The ∆ H o for the reaction C graphite + 1 2 O 2 g → CO g is

Step-by-step solution

(A) 2 CO g + O 2 g → 2 CO 2 g   ΔH 1 o = - x   kJ   mol - 1 (B) C graphite + O 2 g → CO 2 g   ΔH 2 o = - y   kJ   mol - 1 If the equation is multiplied the value n , then the enthalpy change value also multiplied by factor n . If the enthalpy is positive in one direction, it is negative in other direction. Multiply equation B by 2 and subtract equation A from it 2 C graphite + O 2 g → 2 CO g   ΔH ∘ = x - 2 y C graphite + 1 2 O 2 → CO ;