The H _ ( f ) ^ o of NO ₂( ~g ) and N ₂ O ₄( ~g ) are 16.0 and 4.0 k cal mol ⁻¹ respectively. The heat of dimerisation of NO ₂ in k cal is :

Step-by-step solution

The reaction for the dimerisation of NO ₂ is given by: 2 NO ₂( g ) N ₂ O ₄( g ) The enthalpy of reaction H^ _ r is calculated using the standard enthalpies of formation H^ _ f of the products and reactants: H^ _ r = H^ _ f ( products ) - H^ _ f ( reactants ) Given H^ _ f ( NO ₂) = 16.0 kcal mol ⁻¹ and H^ _ f ( N ₂ O ₄) = 4.0 kcal mol ⁻¹ , we substitute these values into the equation: H^ _ r = H^ _ f ( N ₂ O ₄) - 2 H^ _ f ( NO ₂) H^ _ r = 4.0 - 2 16.0 H^ _ r = 4.0 - 32.0 H^ _ r = -28.0 kcal Answer: -28 k cal