Question

The standard enthalpy of formation (\(\Delta H_f^\circ\)) of ammonia is -46.2 kJ/mol. What is the \(\Delta H_f^\circ\) of the following reaction? \[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

• A. -46.2 kJ
• B. +46.2 kJ
• C. -92.4 kJ
• D. -184.8 kJ

Hint:

When using Hess's Law, remember that the enthalpy change for a reaction is the sum of the enthalpy changes for the formation of products and reactants.

Answer 1

The Correct Option is C

For the reaction provided, the change in enthalpy can be calculated using Hess's Law. Given that the formation enthalpy of ammonia is -46.2 kJ/mol, and the reaction produces 2 moles of ammonia, the enthalpy change for the reaction is: \[ \Delta H_f^\circ = 2 \times (-46.2) = -92.4 \, \text{kJ/mol} \] Thus, the standard enthalpy of formation for this reaction is -92.4 kJ.