Question

The enthalpy of formation of ammonia is -46.0koj mol-1 the enthalpy change for the reaction 2nh3 (g)2n2 (g)+3h2(g) is

Answer 1

Chemical Reaction: 2NH3(g) → N2(g) + 3H2(g)

∆Hreaction=2×∆Hof(NH3)-[∆Hof(N2)+3∆Hof(H2)]

∆Hreaction=2(−46)−0=−92 kJ for n2

then 2n2= 2×-92= -184 kj

Answer 2

Answer: 92 kJ/mol

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

$\frac{1}{2}N_2(g)+\frac{3}{2}H_2 (g)\rightarrow NH_3(g)$ $\Delta H=-46.0kJ/mol$

Reversing the reaction, changes the sign of $\Delta H$

$NH_3(g)\rightarrow \frac{1}{2}N_2(g)+\frac{3}{2}H_2(g)$ $\Delta H=46.0kJ/mol$

On multiplying the reaction by 2, enthalpy gets twice:

$2NH_3(g)\rightarrow N_2(g)+3H_2(g)$ $\Delta H=2\times 46.0kJ/mol=92kJ/mol$

Thus the enthalpy change for the reaction $2NH_3(g)\rightarrow N_2(g)+3H_2(g)$ is 92 kJ/mol