Question

What is the heat of reaction (ΔH°rxn) for the combustion of acetone (C3H6O) given the following thermochemical equations? 1. 3 C(s) + 3 H2(g) + ½ O2(g) → C3H6O(ℓ) ΔHf° = −285.0 kJ 2. C(s) + O2(g) → CO2(g) ΔHf° = −394.0 kJ 3. H2(g) + ½ O2(g) → H2O(ℓ) ΔHf° = −286.0 kJ

Answer 1

Answer : The enthalpy of combustion of $C_3H_6O$ will be -1775 kJ

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 combustion of $C_3H_6O$ will be,

$C_3H_6O(l)+4O_2(g)\rightarrow 3CO_2(g)+3H_2O(l)$    $\Delta H_{comb}=?$

The intermediate balanced chemical reaction will be,

(1) $3C(s)+3H_2(g)+\frac{1}{2}O_2(g)\rightarrow C_3H_6O(l)$     $\Delta H_1=-285.0kJ$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$    $\Delta H_2=-394.0kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$    $\Delta H_3=-286.0kJ$

Now we are reversing the reaction 1, multiplying reaction 2 and 3 by 3 and then adding all the equations, we get :

(1) $C_3H_6O(l)\rightarrow 3C(s)+3H_2(g)+\frac{1}{2}O_2(g)$     $\Delta H_1=285.0kJ$

(2) $3C(s)+3O_2(g)\rightarrow 3CO_2(g)$    $\Delta H_2=3\times -394.0kJ=-1182.0kJ$

(3) $3H_2(g)+\frac{3}{2}O_2(g)\rightarrow 3H_2O(l)$    $\Delta H_3=3\times -286.0kJ=-858.0kJ$

The expression for enthalpy of combustion of $C_3H_6O$ will be,

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

$\Delta H_{comb}=(285.0)+(-1182.0)+(-858.0)$

$\Delta H_{comb}=-1755kJ$

Therefore, the enthalpy of combustion of $C_3H_6O$ will be -1775 kJ