Question

A scientist measures the standard enthalpy change for the following reaction to be -190.0 kJ : HCl(g) + NH3(g) NH4Cl(s) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of NH4Cl(s) is kJ/mol.

Answer 1

Answer: The standard enthalpy of formation of $NH_4Cl(s)$ is -328.4 kJ/mol

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f_{(products)}]-\sum [n\times \Delta H_f_{(reactants)}]$

For the given chemical reaction:

$HCl(g)+NH_3(g)\rightarrow NH_4Cl(s)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_f_{(NH_4Cl(s))})]-[(1\times \Delta H_f_{(NH_3(g))})+(1\times \Delta H_f_{(HCl(g))})]$

We are given:

$\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(NH_3)}=-46.1kJ/mol\\\Delta H_{rxn}=-190.0kJ$

Putting values in above equation, we get:

$-190.0=[(1\times \Delta H_f_{(NH_4Cl(s))})]-[(1\times (-46.1))+(1\times (-92.30))]\\\\\Delta H_f_{(NH_4Cl(s))}=-328.4kJ/mol$

Hence, the standard enthalpy of formation of $NH_4Cl(s)$ is -328.4 kJ/mol