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

Question

20 days ago

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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

Chemistry

1 answer:

KiRa [2.7K] · Answer 1 · 2026-03-16T03:59:16+03:00

Answer:

B. –99 kJ.

Explanation:

The following data is provided:

1. C(s) + O₂(g) → CO₂(g);

ΔH = -393 kJ

2. 2CO(g) + O₂ → 2CO₂(g);

ΔH = -588 kJ

Applying Hess's Law, since our desired equation starts with C(s), we can restate equation 1:

1. C(s) + O₂(g) → CO₂(g);

ΔH = -393 kJ

Next, we invert equation 2 and divide it by 2 to obtain equation 3:

3. CO₂(g) → CO(g) + ½O₂;

ΔH = +294 kJ

This means we have switched the sign of ΔH and divided the value by 2. We then combine equations 1 and 3 along with their respective ΔH values.

The result is:

C(s) + ½O₂(g) → CO(g);

ΔH = +294 - 393 kJ

= -99 kJ

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