Question

Given the following at 25°C, calculate ="\Delta H_f" align="absmiddle" class="latex-formula"> for HCN(g) (in kJ/mol) at 25°C.
$2NH_3(g) + 3O_2(g) + 2CH_4(g) \longrightarrow 2HCN(g) + 6H_2O(g)$; ΔH= –870.8 kJ
ΔH = –80.3 kJ/mol for NH₃(g), - 74.6 kJ/mol for CH₄, and - 241.8 kJ/mol for H₂O(g).

Answer 1

Answer: The enthalpy of formation of HCN(g) is 135.1 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:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)$

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

$\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN(g))})+(6\times \Delta H_f_{(H_2O(g))})]-[(2\times \Delta H_f_{(NH_3(g))})+(3\times \Delta H_f_{(O_2(g))})+(2\times \Delta H_f_{(CH_4(g))})]$

We are given:

$\Delta H_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-80.3kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.6kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ$

Putting values in above equation, we get:

$-870.8=[(2\times \Delta H_f_{(HCN(g))})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN(g))}=135.1kJ/mol$

Hence, the enthalpy of formation of HCN(g) is 135.1 kJ/mol