Chapter 9, Problem 65E

Interpretation Introduction

Interpretation: The $\Delta \text{H}$ value for the following reaction needs to be determined:

  $Ca{C}_2(s)+2H_{2}O(l)\to Ca{(OH)}_2(aq)+C_2{H}_2(g)$

Concept Introduction: For the overall processes, the enthalpy change can be determined by adding the enthalpy change of all the steps involved in the process is known as Hess’s Law. The equation to show Hess’s law is:

  ${\text{ΔH}}_{\text{total}}{\text{= ΔH}}_{\text{1}}{\text{+ΔH}}_{\text{2}}{\text{+ΔH}}_{\text{3}}\text{+}\mathrm{...}{\text{+ΔH}}_{\text{n}}$

Answer to Problem 65E

  $\Delta \text{H = }-712.8\text{ kJ}$

Explanation of Solution

Given:

  $\begin{array}{l}Ca(s)+2C(graphite)\to Ca{C}_2(s)\Delta \text{H=-62}\text{.8 kJ -(1)}\\ \text{ Ca(s)+}\frac{1}{2}{O}_2(g)\to CaO(s)\Delta \text{H=}-635.5\text{ kJ -(2)}\\ {\text{ CaO(s)+H}}_{2}O(l)\to Ca{(}_O(aq)\Delta \text{H=}-653.1\text{ kJ -(3)}\\ {\text{ C}}_2{H}_2(g)+\frac{5}{2}{O}_2(g)\to 2C{O}_2(g)+H_{2}O(l)\Delta \text{H=-1300 kJ -(4)}\\ {\text{ C(graphite)+O}}_2\text{(g)}\to {\text{CO}}_2\text{(g) }\Delta \text{H=-}393.5\text{ kJ -(5)}\end{array}$

Rules related to enthalpies of the reaction are:

• When a reaction is inverted, then the sign of enthalpy is also inverted.
• When a reaction is multiplied by ‘n’ coefficient, then the value of enthalpy is multiplied by ‘n’ value.

The required equation is:

  $Ca{C}_2(s)+2H_{2}O(l)\to Ca{(OH)}_2(aq)+C_2{H}_2(g)$

In order to obtain the required equation, reversing reaction (1) and adding to reaction (3).

  $\begin{array}{l}Ca{C}_2(s)\to Ca(s)+2C(graphite)\Delta \text{H=+62}\text{.8 kJ}\\ \underset{\_}{\text{CaO(s)}+{\text{H}}_{2}O(l)\to Ca{(OH)}_2(aq)}\Delta \text{H=}-653.1\text{ kJ}\\ Ca{C}_2(s)+\text{CaO(s)}+{\text{H}}_{2}O(l)\to Ca(s)+2C+Ca{(}_O(aq)\Delta \text{H=-590}\text{.3 kJ -(6) }\\ \end{array}$

Adding reaction (2) to reaction (6)

  $\begin{array}{l}Ca{C}_2(s)+\text{CaO(s)}+{\text{H}}_{2}O(l)\to Ca(s)+2C+Ca{(}_O(aq)\Delta \text{H=-590}\text{.3 kJ -(6) }\\ \underset{\_}{\text{ Ca(s)+}\frac{1}{2}{O}_2(g)\to CaO(s)\Delta \text{H=}-635.5\text{ kJ -(2)}}\\ Ca{C}_2(s)+\text{CaO(s)}+{\text{H}}_{2}O(l)+\text{Ca(s)+}\frac{1}{2}{O}_2(g)\to Ca(s)+2C+Ca{(}_O(aq)+CaO(s)\Delta \text{H=}-1225.8\text{ kJ -(7) }\end{array}$

Now multiply reaction (5) by 2 and add reaction (7) to it:

  $\begin{array}{l}{\text{2C(graphite)+2O}}_2\text{(g)}\to 2{\text{CO}}_2\text{(g) }\Delta \text{H=-787 kJ }\\ \underset{\_}{Ca{C}_2(s)+\text{CaO(s)}+{\text{H}}_{2}O(l)+\text{Ca(s)+}\frac{1}{2}{O}_2(g)\to Ca(s)+2C+Ca{(OH)}_2(aq)+CaO(s)\Delta \text{H=}-1225.8\text{ kJ }}\\ Ca{C}_2(s)+{\text{H}}_{2}O(l)+\frac{5}{2}{\text{O}}_2\text{(g)}\to 2{\text{CO}}_2\text{(g)}+Ca{(}_O(aq)\Delta \text{H=}-2012.8\text{ kJ -(8)}\end{array}$

Now reverse the reaction (4) and add to the reaction (8).

  $\begin{array}{l}2C{O}_2(g)+H_{2}O(l)\to {\text{C}}_2{H}_2(g)+\frac{5}{2}{O}_2(g)\Delta \text{H=+1300 kJ }\\ \underset{\_}{Ca{C}_2(s)+{\text{H}}_{2}O(l)+\frac{5}{2}{\text{O}}_2\text{(g)}\to 2{\text{CO}}_2\text{(g)}+Ca{(OH)}_2(aq)\Delta \text{H=}-2012.8\text{ kJ }}\\ 2C{O}_2(g)+2H_{2}O(l)+\frac{5}{2}{\text{O}}_2\text{(g)+}Ca{C}_2(s)\to {\text{C}}_2{H}_2(g)+\frac{5}{2}{O}_2(g)+2{\text{CO}}_2\text{(g)}+Ca{(}_O(aq)\Delta \text{H=}-712.8\text{ kJ}\end{array}$

Cancelling the common terms in above equation:

  $Ca{C}_2(s)+2H_{2}O(l)\to Ca{(OH)}_2(aq)+C_2{H}_2(g)\Delta \text{H = }-712.8\text{ kJ}$