Chapter 21, Problem 97A

Interpretation Introduction

Interpretation: The amount of heat liberated when 4.80 g of methane reacts with an excess of oxygen needs to be determined.

Concept Introduction: The relation between number of moles, mass and molar mass is as follows:

  $n=\frac{m}{M}$

Here, m is mass, M is molar mass and n is number of moles.

Explanation of Solution

The combustion of methane is represented follows:

  ${\text{CH}}_{\text{4}}\left(g\right)+2{\text{O}}_{\text{2}}\left(g\right)\to {\mathrm{CO}}_2\left(g\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)$

The change in enthalpy of reaction is -890 kJ. Since the change in enthalpy is negative, the reaction is exothermic in nature.

In the above reaction, number of moles of methane is 1 mol. Thus, for 1 mol -890 kJ of heat is liberated.

The given amount of methane is 4.80 g in the excess of oxygen.

The molar mass of methane is 16.04 g/mol thus, number of moles can be calculated as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Substitute the values,

  $\begin{array}{c}n=\frac{4.80\text{ g}}{16.04\text{ g/mol}}\\ =0.299\text{ mol}\end{array}$

For 1 mol of methane, the amount of heat liberated is -890 kJ thus, for 0.299 mol, the amount of heat liberated will be:

  $\begin{array}{c}\Delta H=\left(0.299\text{ mol}\right)\left(\frac{-890\text{ kJ}}{1\text{ mol}}\right)\\ =-266.33\text{ kJ}\end{array}$

Thus, the amount of heat liberated is $-266.33\text{ kJ}$ .