Question

The standard enthalpy of formation of ammonia is −46.0 kJ mol⁻¹. The enthalpy change for the reaction:

\[\ce{2NH3_{(g)} -> 2N2_{(g)} + 3H2_{(g)}}\] is ____________.

विकल्प

• 46.0 kJ mol⁻¹

• 92.0 kJ mol⁻¹

• −23.0 kJ mol⁻¹

• −92.0 kJ mol⁻¹

Answer 1

The standard enthalpy of formation of ammonia is −46.0 kJ mol⁻¹. The enthalpy change for the reaction:

\[\ce{2NH3_{(g)} -> 2N2_{(g)} + 3H2_{(g)}}\] is 92.0 kJ mol⁻¹.

Explanation:

Because the formation of 1 mole of ammonia releases −46.0 kJ mol⁻¹ of energy, the breakdown of 1 mole of ammonia consumes the same amount of energy, i.e. +46.0 kJ mol⁻¹. The energy spent will be +92.0 kJ mol⁻¹ since the described reaction implies the breakdown of two moles of ammonia.