Question

The standard molar enthalpy of formation of NH3(g) is -45.9 kJ/mol. What is the enthalpy change if 9.51 g N2(g) and 1.96 g H2(g) react to produce NH3(g)

Answer 1

Answer:

$\Delta H=-29.7kJ$

Explanation:

Hello!

In this case, since the undergoing chemical reaction is:

$N_2+3H_2\rightarrow 2NH_3$

We first need to identify the limiting reactant given the masses of nitrogen and hydrogen:

$n_{NH_3}^{by\ H_2}=1.96gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.647molNH_3\\\\ n_{NH_3}^{by\ N_2}=9.51gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.679molNH_3$

It means that only 0.647 moles of ammonia are yielded, so the resulting enthalpy change is:

$\Delta H=0.647molNH_3*\frac{-45.9kJ}{1molNH_3}\\\\ \Delta H=-29.7kJ$

Best regards!