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).