Enthalpy of formation of CO (g), CO ₂(g), N ₂ O (g) and N ₂ O ₄(g) are -110,-393,81,9.7 ~kJ ~mol ⁻¹ respectively. Calculate _r H for the following reaction. N ₂ O ₄(g)+3 CQ (g) N ₂ O (g)+3 CO ₂(g)

Step-by-step solution

Given, Enthalpy of formation of CO (g)=-110 ~kJ ~mol ⁻¹ Enthalpy of formation of CO ₂(g)=-393 ~kJ ~mol ⁻¹ Enthalpy of formation of N ₂ O (g)=81 ~kJ ~mol ⁻¹ Enthalpy of formation of N ₂ O ₄(g)=9.7 ~kJ ~mol ⁻¹ N ₂ O ₄(g)+3 CO (g) N ₂ O (g)+3 CO ₂(g) aligned _r H & = H_ Product - H_ Reactant & =[(81)+3(-393)]-[(9.7)+3(-110)] & =[81-1179]-[9.7-330] & =(-1098)-(-320.3)=-777.7 & -778 ~kJ ~mol ⁻¹ aligned