Question

CO(g) + 12 O2(g) → CO2(g)The combustion of carbon monoxide is represented by the equation above.(a) Determine the value of the standard enthalpy change, ∆HrxnD , for the combustion of CO(g) at 298 Kusing the following information. C(s) + 12 O2(g) → CO(g) ∆H298D = − 110.5 kJ mol−1 C(s) + O2(g) → CO2(g) ∆H298D = − 393.5 kJ mol−1

Answer 1

Answer : The standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol

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 $CO$ will be,

$CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)$    $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $C(s)+\frac{1}{2}O_2(g)\rightarrow CO(g)$     $\Delta H_1=-110.5kJ/mol$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$     $\Delta H_2=-393.5kJ/mol$

Now we are reversing reaction 1 and then adding both the equations, we get :

(1) $CO(g)\rightarrow C(s)+\frac{1}{2}O_2(g)$     $\Delta H_1=110.5kJ/mol$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$     $\Delta H_2=-393.5kJ/mol$

The expression for enthalpy change for the reaction will be,

$\Delta H_{rxn}=\Delta H_1+\Delta H_2$

$\Delta H_{rxn}=(110.5)+(-393.5)$

$\Delta H_{rxn}=-283kJ/mol$

Therefore, the standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol