Question

Use the following information S(s)+O2(g)→SO2(g), ΔH∘ = -296.8kJ SO2(g)+12O2(g)→SO3(g) , ΔH∘ = -98.9kJ to calculate ΔH∘f for H2SO4(aq) (in kilojoules per mole). [For H2O(l),ΔH∘f = -285.8kJ/mol].

Answer 1

Answer : The enthalpy of formation of $H_2SO_4$ is, -812.4 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation of $H_2SO_4$ will be,

$S+H_2+2O_2\rightarrow H_2SO_4$    $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $S+O_2\rightarrow SO_2$     $\Delta H_1=-298.2$

(2) $SO_2+\frac{1}{2}O_2\rightarrow SO_3$    $\Delta H_2=-98.2$

(3) $SO_3+H_2O\rightarrow H_2SO_4$    $\Delta H_3=-130.2$

(4) $H_2+\frac{1}{2}O_2\rightarrow H_2O$    $\Delta H_4=-285.8$

Now adding all the equations, we get the expression for enthalpy of formation of $H_2SO_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4$

$\Delta H=(-298.2)+(-98.2)+(-130.2)+(-285.8)$

$\Delta H=-812.4kJ/mol$

Therefore, the enthalpy of formation of $H_2SO_4$ is, -812.4 kJ/mole