Question

question 3 (1 point)
the enthalpy change for reaction $n_2 + 3h_2(g) \to 2nh_3(g)$ is 92.2kj. the enthalpy of formation of ammonia is
-92.2 kj
-46.1 kj
-46.1 kj/mol
-92.2 kj/mol
cannot be determined

Explanation:

Step1: Recall the definition of enthalpy of formation

The enthalpy of formation ($\Delta H_f$) is the enthalpy change when one mole of a substance is formed from its elements in their standard states.
For the reaction \(N_2 + 3H_2(g)\to2NH_3(g)\), \(\Delta H=- 92.2\space kJ\) (since it is an exothermic reaction, the value is negative).

Step2: Calculate the enthalpy of formation per mole of \(NH_3\)

If \(2\) moles of \(NH_3\) have an enthalpy change of \(-92.2\space kJ\), then for \(1\) mole of \(NH_3\):

$$ \Delta H_f=\frac{-92.2\space kJ}{2}=- 46.1\space kJ/mol $$

Answer:

C. \(-46.1\space kJ/mol\)