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

`\Delta H=-29.7kJ`

Step-by-step 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*(1molH_2)/(2.02gH_2)*(2molNH_3)/(3molH_2)=0.647molNH_3\\\\ n_(NH_3)^(by\ N_2)=9.51gN_2*(1molN_2)/(28.02gN_2)*(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*(-45.9kJ)/(1molNH_3)\\\\ \Delta H=-29.7kJ`

Best regards!