Chemistry - Thermochemistry Question with Solution | TestHub

To find the heat of formation of methane, C(s) + 2H₂(g) $⟶$ CH₄(g), we apply Hess's Law.

Given reactions:

1. C(s) + O₂(g) $⟶$ CO₂(g); $\Delta H_1=-94.2$ kcal mol⁻¹

2. H₂(g) + ½O₂(g) $⟶$ H₂O(l); $\Delta H_2=-68.3$ kcal mol⁻¹

3. CH₄(g) + 2O₂(g) $⟶$ CO₂(g) + 2H₂O(l); $\Delta H_3=-210.8$ kcal mol⁻¹

We need to manipulate these equations to get the target equation.

Keep reaction (1) as is:

C(s) + O₂(g) $⟶$ CO₂(g); $\Delta H_1=-94.2$ kcal mol⁻¹

Multiply reaction (2) by 2:

2H₂(g) + O₂(g) $⟶$ 2H₂O(l); $2\times \Delta H_2=2\times (-68.3)=-136.6$ kcal mol⁻¹

Reverse reaction (3):

CO₂(g) + 2H₂O(l) $⟶$ CH₄(g) + 2O₂(g); $-\Delta H_3=-(-210.8)=+210.8$ kcal mol⁻¹

Add the manipulated equations:

C(s) + O₂(g) + 2H₂(g) + O₂(g) + CO₂(g) + 2H₂O(l) $⟶$ CO₂(g) + 2H₂O(l) + CH₄(g) + 2O₂(g)

Cancel common species on both sides:

C(s) + 2H₂(g) $⟶$ CH₄(g)

The enthalpy of formation of methane is the sum of the manipulated $\Delta H$ values:

$\Delta H_f$(CH₄) = $\Delta H_1+(2\times \Delta H_2)-\Delta H_3$

$\Delta H_f$(CH₄) = $-94.2+(2\times -68.3)-(-210.8)$

$\Delta H_f$(CH₄) = $-94.2-136.6+210.8$

$\Delta H_f$(CH₄) = $-230.8+210.8$

$\Delta H_f$(CH₄) = $-20.0$ kcal mol⁻¹

The final answer is $\boxed{-20.0kcalmo{l}^{-1}}$.