Chapter 17, Problem 82A

Interpretation Introduction

Interpretation: The amount of heat liberated during the chemical reaction is to be found.

Concept introduction: The reactant in a chemical process known as the limiting reactant controls how much product can be produced. When the limiting reactant is fully consumed, the reaction will come to an end. The reactant in a chemical reaction known as the excess reactant is what is left over after the limiting reactant has been entirely consumed.

Answer to Problem 82A

The amount of heat liberated when $4.79\text{ g}$ of ${\text{C}}_2{\text{H}}_4$ reacts with excess oxygen is $238\text{ kJ}$

Explanation of Solution

The given chemical reaction is

  ${\text{C}}_2{\text{H}}_4{\text{(g)+3O}}_{\text{2}}\text{(g)}\to 2{\text{CO}}_{\text{2}}{\text{(g)+2H}}_{\text{2}}\text{O(l) }\Delta \text{H= -1}\text{.40}\times {\text{10}}^3\text{ kJ}$

The enthalpy change is negative which means the reaction is exothermic. Also, it is given that oxygen is in excess, thus ${\text{C}}_2{\text{H}}_4$ is the limiting reactant and will decide the amount of formed and heat liberated.

Given amount of ${\text{C}}_2{\text{H}}_4$ = $4.79\text{ g}$

  $\begin{array}{l}\text{moles =}\frac{\text{given amount}}{\text{Molecular Mass}}\\ {\text{moles of C}}_2{\text{H}}_4=\frac{\text{4}\text{.79 g}}{\text{28}\text{.05 g/mol}}=0.170\text{ mol}\end{array}$

It is given that when $1\text{ mol}$ of ${\text{C}}_2{\text{H}}_4$ reacts with oxygen, the energy liberated is $\text{1}\text{.40}\times {\text{10}}^3\text{ kJ}$

Thus, when $0.170\text{ mol}$ of ${\text{C}}_2{\text{H}}_4$ reacts with oxygen, the energy liberated is = $\frac{\text{1}\text{.40}\times {\text{10}}^3\text{ kJ}}{\text{1 mol}}\times 0.170\text{ mol}=238\text{ kJ}$