Question

The enthalpy change for the reaction
C(𝑔) +$\frac{1}{2}$ O₂(𝑔)→CO(𝑔) is _______ kJ per mole of CO(𝑔) produced. 
(rounded off to one decimal place) 
[Given: 
C(𝑔)+O₂(𝑔) → CO₂(𝑔), ΔH_rxn = −393.5 kJ per mole of CO₂(𝑔) produced 
CO₂(𝑔) → CO(𝑔) +$\frac{1}{2}$ O₂(𝑔), ΔH_rxn = 283.0 kJ per mole of CO(𝑔) produced]

Answer 1

Correct Answer: -111

We are given the following reactions and their enthalpy changes:

$C(g) + O_2(g) \rightarrow CO_2(g), \quad \Delta H_{rxn} = -393.5 \text{ kJ per mole of } CO_2(g).$

$CO_2(g) \rightarrow CO(g) + \frac{1}{2} O_2(g), \quad \Delta H_{rxn} = 283.0 \text{ kJ per mole of } CO(g).$

To calculate the enthalpy change for the reaction:

$C(g) + \frac{1}{2} O_2(g) \rightarrow CO(g),$

we need to manipulate the given reactions. First, reverse the second reaction so that the products and reactants match. This reverses the sign of the enthalpy change for that reaction.

Reversing $CO_2(g) \rightarrow CO(g) + \frac{1}{2}O_2(g)$ gives:

$CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g), \quad \Delta H_{rxn} = -283.0 \text{ kJ per mole of } CO(g).$

Now add the two reactions:

$C(g) + O_2(g) \rightarrow CO_2(g), \quad \Delta H_{rxn} = -393.5 \text{ kJ per mole of } CO_2(g),$

$CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g), \quad \Delta H_{rxn} = -283.0 \text{ kJ per mole of } CO(g).$

By adding these reactions, the intermediate $CO_2(g)$ cancels out, giving the desired reaction.

The total enthalpy change is:

$\Delta H_{rxn} = -393.5 + 283.0 = -110.5 \text{ kJ per mole of } CO(g).$

Thus, the enthalpy change is approximately -110.5 kJ per mole of CO(g).