Question

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

Answer 1

Answer:

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.