Chapter 6, Problem 6.151QP

Graphite is burned in oxygen to give carbon monoxide and carbon dioxide. If the product mixture is 33% CO and 67% CO₂ by mass, what is the heat from the combustion of 1.00 g of graphite?

Interpretation Introduction

Interpretation:

The amount of heat evolved from the combustion of $\text{1}\text{g}\text{of}$ graphite has to be calculated.

Answer to Problem 6.151QP

• The amount of heat evolved from the given combustion reaction is $\text{22}\text{kJ}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}\text{Graphite}\text{is}\text{burned}\text{in}\text{oxygen and}\text{gives CO}\text{and}{\text{CO}}_{\text{2}}\text{.}\\ \text{product}\text{mixture}\text{contains}\text{33\%}\text{CO}\text{and}\text{67\%}{\text{CO}}_{\text{2}}\text{.}\end{array}$

Calculate the mole fraction of each gas present in the product, assuming $\text{100}\text{g}$ product:

$\begin{array}{l}\text{Mol}\text{CO}\text{=}\text{33}\text{g}\text{CO}\text{×}\text{1}\text{mol}\text{CO/28}\text{.01}\text{g}\text{CO =}\text{1}\text{.178}\text{mol}\text{CO}\\ \text{Mol}{\text{CO}}_{\text{2}}\text{=}\text{67}\text{g}{\text{CO}}_{\text{2}}\text{×}\text{1}\text{mol}{\text{CO}}_{\text{2}}\text{/44}\text{.01}\text{g}{\text{CO}}_{\text{2}}\text{ =}\text{1}\text{.522}\text{mol}{\text{CO}}_{\text{2}}\\ \text{Mol}\text{fraction}\text{CO}\text{=}\frac{\text{1}\text{.178}\text{mol}\text{CO}}{\left(\text{1}\text{.178}\text{+}\text{1}\text{.522}\right)\text{mol}}\text{=}\text{0}\text{.4363}\\ \text{Mol}\text{fraction}{\text{CO}}_{\text{2}}\text{=}\frac{\text{1}\text{.522}\text{mol}{\text{CO}}_{\text{2}}}{\left(\text{1}\text{.178}\text{+}\text{1}\text{.522}\right)\text{mol}}\text{=}\text{0}\text{.5637}\end{array}$

Calculate the starting moles of $\text{C}$, which gives the total moles of $\text{CO}\text{and}{\text{CO}}_{\text{2}}$

$\begin{array}{l}\text{Starting}\text{mol}{\text{C}}_{\left(\text{s}\right)}\text{=}\text{1}\text{.00}\text{g}\text{C}\text{×}\text{1}\text{mol}\text{C/12}\text{.01}\text{g}\text{C}\\ \text{=}\text{0}\text{.08326}\text{mol}\text{C}\text{=}\text{mol}\text{CO}\text{+}{\text{CO}}_{\text{2}}\end{array}$

Calculate the moles of $\text{CO}\text{and}{\text{CO}}_{\text{2}}$ using mole fraction values

$\begin{array}{l}\text{Mol}\text{CO}\text{=}\frac{\text{0}\text{.4363}\text{mol}\text{CO}}{\text{1}\text{mol}\text{total}}\text{×}\text{0}\text{.08326}\text{mol}\text{total}\text{=}\text{0}\text{.03633}\text{mol}\text{CO}\\ \text{Mol}{\text{CO}}_{\text{2}}\text{=}\frac{\text{0}\text{.5637}\text{mol}{\text{CO}}_{\text{2}}}{\text{1}\text{mol}\text{total}}\text{×}\text{0}\text{.08326}\text{mol}\text{total}\text{=}\text{0}\text{.04693}\text{mol}{\text{CO}}_{\text{2}}\end{array}$

Calculate the heat of combustion by using enthalpies of formation

$\begin{array}{l}{\text{C}}_{\left(\text{s}\right)}\text{+}{\text{1/2O}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{CO}}_{\left(\text{g}\right)}\\ \text{0}\text{.03633}\text{excess}\text{0}\text{.03633}\text{mol}\\ \text{0}\text{0}\text{-}\text{110}\text{.5}\text{kJ/mol}\\ \text{0}\text{0}\text{-}\text{4}\text{.014}\text{kJ/0}\text{.03633}\text{mol}\\ \\ {\text{C}}_{\left(\text{s}\right)}\text{+}{\text{O}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{CO}}_{\text{2}}{}_{\left(\text{g}\right)}\\ \text{0}\text{.04693}\text{excess}\text{0}\text{.04693}\text{mol}\\ \text{0}\text{0}\text{-393}\text{.5}\text{kJ/mol}\\ \text{0}\text{0}\text{-}\text{18}\text{.47}\text{kJ/0}\text{.04693}\text{mol}\end{array}$

Calculate the total enthalpy change

$\begin{array}{l}\text{total}\text{ΔH}\text{=}\text{-}\text{4}\text{.014}\text{+}\left(\text{-}\text{18}\text{.47}\right)\\ \text{=}\text{-}\text{22}\text{.48}\text{=}\text{-}\text{22}\text{kJ}\text{.}\end{array}$

The amount of heat evolved from the given combustion reaction is $\text{22}\text{kJ}\text{.}$

Conclusion

The amount of heat evolved from the given reaction was calculated by using mole and enthalpy change calculations.