Chapter 19, Problem 19.54QA

Interpretation Introduction

To calculate:

a) $∆H_{rxn}$ for the production of acetylene from the reaction between calcium carbide and water.

b) Is the given reaction endothermic or an exothermic?

Answer to Problem 19.54QA

Solution:

a) $∆H_{rxn}$ for the production of acetylene from the reaction between calcium carbide and water is  $-127.1 kJ$.

b) The reaction is exothermic.

Explanation of Solution

1) Concept:

We can calculate the $∆H_{rxn}$ from the difference between the standard molar enthalpies of reactants and products.

2) Formula:

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=\sum {\mathrm{n}}_{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{s}}\times ∆{\mathrm{H}}_{\mathrm{f},\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{s}}^0-\sum {\mathrm{n}}_{\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{s}}\times ∆{\mathrm{H}}_{\mathrm{f},\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{s}}^0$

“n” is the coefficient of the products and reactants.

3) Given:

The given reaction is

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

From appendix 4, the value of $∆{\mathrm{H}}_{\mathrm{f}}^0$ are

	$∆{\mathrm{H}}_{\mathrm{f}}^0(\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l})$
$Ca{C}_2\left(s\right)$	$-59.8$
$H_{2}O \left(l\right)$	$-285.8$
$C_2{H}_2\left(g\right)$	$226.7$
$Ca{\left(OH\right)}_2\left(s\right)$	$-985.2$

4) Calculation:

a) Calculation for $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0$:

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=\sum {\mathrm{n}}_{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{s}}\times ∆{\mathrm{H}}_{\mathrm{f},\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{s}}^0-\sum {\mathrm{n}}_{\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{s}}\times ∆{\mathrm{H}}_{\mathrm{f},\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{s}}^0$

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=\left[\left(1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times ∆{\mathrm{H}}_{\mathrm{f},C_2{H}_2\left(g\right)}^0\right)+\left(1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times ∆{\mathrm{H}}_{\mathrm{f},Ca{\left(OH\right)}_2\left(s\right)}^0\right)\right]-\left[\left(1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times ∆{\mathrm{H}}_{\mathrm{f},Ca{C}_2\left(s\right)}^0\right)+\left(2\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times ∆{\mathrm{H}}_{\mathrm{f},H_{2}O \left(l\right)}^0\right)\right]$

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=\left[\left(1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times 226.7\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)+\left(1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times \left(-985.2\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)\right)\right]-\left[\left(1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times \left(-59.8\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)\right)+\left(2\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times -285.8\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)\right]$

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=\left(-758.5\right)-(-631.4)$

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=-127.1 kJ$

$∆H_{rxn}$ for the production of acetylene from the reaction between calcium carbide and water is $-127.1 kJ$.

b) The sign of calculated value $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}$ is negative. This indicates that the reaction is exothermic.

Conclusion:

Based on the values of $∆H_f^0$, the value of $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}$ for the production of acetylene from the reaction between calcium carbide and water is calculated, and the reaction is exothermic.