Question

Given the following at 25C calculate delta Hf for HCN (g) at 25C. 2NH3 (g) +3O2 (g) + 2CH4 (g) ---> 2HCN (g) + 6H2O (g) delta H rxn= -870.8 kJ. delta Hf=-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 $\Delta H_f$ for HCN (g) in the reaction 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:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

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:

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

We are given:

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

Putting values in above equation, we get:

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

Hence, the $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

Answer 2

The heat of formation is the heat reactants have or uses to form the products during a chemical reaction. The heat of formation $(\rm \Delta H_{f})$ is 135.1 kJ/mol.

What is the heat of formation?

Given,

• The heat of formation $(\rm \Delta H_{f})$ of water = -241.8 kJ/mol
• The heat of formation $(\rm \Delta H_{f})$ of ammonia = -80.3 kJ/mol
• The heat of formation $(\rm \Delta H_{f})$ of methane = -74.6 kJ/mol
• The heat of formation $(\rm \Delta H_{f})$ = 0 kJ/mol
• Heat change of reaction $(\rm \Delta H_{reaction})$ = -870.8 kJ

The balanced reaction will have 2 moles of ammonia, 3 moles of oxygen and 2 moles of methane while the product will have 2 moles of hydrogen cyanide and 6 moles of water.

Calculate heat change of the reaction by:

ΔH(Reaction) = Σ (n x ΔH(product))-Σ (n x ΔH(reactant))

The equation for the heat change for the reaction is:

-870.8 = [(2 x ΔH(formation) + (6 x (-241.8)] - [ x (-80.3)+ (3 x 0) + (2 x (-74.6)]

ΔH(formation) = 135.1 kJ

Therefore, 135.1 kJ per mol is the heat of formation ΔH(formation).

Learn more about the heat of formation here:

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