Consider the following thermochemical equation: C (s) + O2 (g) → CO2 (g) ΔH = - 393 kJ CO (g) + ½O2 (g) → CO2 (g) ΔH = - 294 kJ What is the enthalpy change for the following related thermochemical equation C (s) + ½O2 (g) → CO (g) Group of answer choices a. - 687 kJ b. - 99 kJ c. + 99 kJ d. + 687 kJ

B. - 99 kJ.

Explanation:

We have the following information:

1. C (s) + O₂ (g) → CO₂ (g);

ΔH = - 393 kJ

2. 2CO (g) + O₂ → 2CO₂ (g);

ΔH = - 588 kJ

Using Hess's Law, Our target equation has C (s) on the left hand side, so we re-write equation 1:

1. C (s) + O₂ (g) → CO₂ (g);

ΔH = - 393 kJ

So, we reverse equation 2 and divide by 2, we have equation 3:

3. CO₂ (g) → CO (g) + ½O₂;

ΔH = + 294 kJ

That is, change the sign of ΔH and divide by 2. Then we add equations 1 and 3 and their ΔH values.

This gives:

C (s) + ½O₂ (g) → CO (g);

ΔH = + 294 - 393 kJ

= - 99 kJ

The standard enthalpy of formation of carbon monoxide is - 99 kJ/mol.