Chapter 5, Problem 135AP

Acetylene $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)$ can be made by combining calcium carbide $\left({\text{CaC}}_{\text{2}}\right)$ with water. (a) Write an equation for the reaction. (b) What is the maximum amount of beat (in Joules) that can be obtained from the combustion of acetylene, starting with $\text{74}{\text{.6g of CaC}}_{\text{2}}$ ?

Interpretation Introduction

Interpretation:

The balanced equation for the given reaction is to be written and the maximum amount of heat that can be obtained from the combustion of acetylene is to be calculated.

Concept introduction:

The standard enthalpy for a reaction is the amount of enthalpy change which occurs at standard conditions.

The standard enthalpy of a reaction is to be determined by using the equation as given below:

$\Delta H{°}_{\text{rxn}}={\displaystyle \sum n}\Delta H_{\text{f}}^{\text{°}}\left(\text{products}\right)-{\displaystyle \sum m}\Delta H_{\text{f}}^{°}\left(\text{reactants}\right)$

Here, the stoichiometric coefficients are represented by m for the reactants and n for the products. The enthalpy of formation at standard conditions is represented by $\Delta H_{\text{f}}^{°}$.

The value of enthalpy of formation of an element is zero at its most stable state.

In one joule, there are $0.001\text{ kJ}$.

Conversion factor for J to kJ is $\frac{\text{0}\text{.001 kJ}}{1\text{ J}}$.

Answer to Problem 135AP

Solution:

(a) ${\text{CaC}}_{\text{2}}\left(s\right)+2{\text{ H}}_{\text{2}}\text{O}\left(l\right)\to \text{Ca}{\left(\text{OH}\right)}_2\left(s\right)+{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)$

(b) $1.51\times {10}^6\text{ J}$

Explanation of Solution

a) An equation for reaction.

The reaction of calcium carbide with water in order to form acetylene is as follows:

${\text{CaC}}_{\text{2}}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{Ca}{\left(\text{OH}\right)}_2\left(s\right)+{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)$

Now, balance this equation to obtain the balanced chemical equation as shown below:

${\text{CaC}}_{\text{2}}\left(s\right)+2{\text{ H}}_{\text{2}}\text{O}\left(l\right)\to \text{Ca}{\left(\text{OH}\right)}_2\left(s\right)+{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)$

Hence, the balanced chemical equation is:

${\text{CaC}}_{\text{2}}\left(s\right)+2{\text{ H}}_{\text{2}}\text{O}\left(l\right)\to \text{Ca}{\left(\text{OH}\right)}_2\left(s\right)+{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)$.

Given information: Mass, $m_{{\text{CaC}}_{\text{2}}}=74.6\text{ g}$.

b) The maximum amount of heat that can be obtained from the combination of acetylene, starting with $74.6\text{ g}$ of ${\text{CaC}}_{\text{2}}$.

The combustion of acetylene takes place as follows:

$2{\text{ C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)+5\text{}{\text{ O}}_{\text{2}}\left(g\right)\to 4{\text{ CO}}_2\left(g\right){\text{+2 H}}_2\text{O}\left(l\right)$

The enthalpy of formation of oxygen gas is zero because it is in the most stable form.

From appendix 2,

$\begin{array}{c}\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)=226.6\text{}\text{kJ}/mol\\ \Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)=-285.8\text{ kJ}/mol\\ \Delta H{°}_{\text{f}}\left({\text{CO}}_{\text{2}}\right)=-393.5\text{ kJ}/mol\end{array}$

Calculate the standard enthalpy of reaction as follows:

$\Delta H{°}_{\text{rxn}}=\left[4\Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)+2\Delta H{°}_{\text{f}}\left({\text{H}}_2\text{O}\right)\right]-\left[2\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{2}}{\text{H}}_2\right)+5\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)\right]$

Substitute $226.6\text{}\text{kJ}/mol$ for $\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\right)$, $-285.8\text{ kJ}/mol$ for $\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)$, $-393.5\text{ kJ}/mol$ for $\Delta H{°}_{\text{f}}\left({\text{CO}}_{\text{2}}\right)$, and 0 for $\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)$ in the above equation and solve:

$\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[4\left(-393.5\text{ kJ}/mol\right)+2\left(-285.8\text{ kJ}/mol\right)\right]-\left[2\left(226.6\text{ kJ}/mol\right)+5\left(0\right)\right]\\ =-2599\text{ kJ}/mol\end{array}$

This enthalpy is required for the formation of $2\text{ mol}$

of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$.

The molar mass of ${\text{CaC}}_{\text{2}}$

is $64\text{ g}/\text{mol}$

Th mass of ${\text{CaC}}_{\text{2}}$ is $74.6\text{ g}$

Calculate the number of moles as follows:

$n_{{\text{CaC}}_{\text{2}}}=\frac{m_{{\text{CaC}}_{\text{2}}}}{M_{{\text{CaC}}_{\text{2}}}}$

Substitute $74.6\text{ g}$ for $m_{{\text{CaC}}_{\text{2}}}$ and $64\text{ g}/\text{mol}$ for $M_{{\text{CaC}}_{\text{2}}}$ in the above equation.

$\begin{array}{c}{n}_{{\text{CaC}}_{\text{2}}}=\frac{74.6\text{ g}}{64\text{ g}/\text{mol}}\\ =1.16\text{ mol}\end{array}$

One mol of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ is produced from one mol of ${\text{CaC}}_{\text{2}}$, and therefore the number of moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$

is equal to number of moles of ${\text{CaC}}_{\text{2}}$.

$n_{{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}}=1.16\text{ mol}$

Calculate the amount of heat required when $1.16\text{ mol}$ of acetylene is burned as follows:

$\left(1.16\cancel{{\text{mol C}}_2{\text{H}}_2}\right)\left(\frac{2599\text{ kJ}}{2\cancel{{\text{ mol C}}_2{\text{H}}_2}}\right)=1.51\times {10}^3\text{ kJ}$

In one kilojoule, there are $1000\text{ J}$.

Convert kilojoule to joule as follows:

$\begin{array}{c}1.51\times {10}^3\text{ kJ}=\left(1.51\times {10}^3\cancel{\text{kJ}}\right)\left(\frac{1000\text{ J}}{1\cancel{\text{kJ}}}\right)\\ =1.51\times {10}^6\text{ J}\end{array}$

Hence, the heat produced from combustion of acetylene is $1.51\times {10}^6\text{ J}$