Chapter 19, Problem 19.55QA

Interpretation Introduction

To estimate:

The $∆H_{rxn}$ for hydrogenation of acetylene ($C_2{H}_2$) with one mole of hydrogen gas to make ethylene ($C_2{H}_4$).

Answer to Problem 19.55QA

Solution:

The $∆H_{rxn}$ for hydrogenation of acetylene ($C_2{H}_2$) with one mole of hydrogen gas to make ethylene ($C_2{H}_4$) is $-174.30 kJ$.

Explanation of Solution

1) Concept:

We can calculate the $∆H_{rxn}$ from the difference between the standard molar enthalpies of reactants and products. We are given two equations with thermochemical data and a third reaction for which we are asked to find $∆H^0$. We can manipulate the equations algebraically and then add to give the equation for which $∆H^0$ is unknown. Then we can calculate the unknown value by applying Hess’s law.

2) Formula:

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

3) Given:

The given reactions are

Equation A: $HC\equiv CH \left(g\right)+2H_2\left(g\right)\to C{H}_{3}C{H}_3\left(g\right)$

Equation B: $H_{2}C=C{H}_2 \left(g\right)+H_2\left(g\right)\to C{H}_{3}C{H}_3\left(g\right)$

Final reaction: $HC\equiv CH \left(g\right)+H_2\left(g\right)\to H_{2}C=C{H}_2 \left(g\right)$

	$∆{\mathrm{H}}_{\mathrm{f}}^0(\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l})$
$HC\equiv CH \left(g\right)$	$226.7$
$H_2\left(g\right)$	$0.0$
$C{H}_{3}C{H}_3\left(g\right)$	$-84.67$
$H_{2}C=C{H}_2 \left(g\right)$	$52.4$

4) Calculation:

Calculation for $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}$ for equation A:

$HC\equiv CH \left(g\right)+2H_2\left(g\right)\to C{H}_{3}C{H}_3\left(g\right)$

	$∆{\mathrm{H}}_{\mathrm{f}}^0(\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l})$
$HC\equiv CH \left(g\right)$	$226.7$
$H_2\left(g\right)$	$0.0$
$C{H}_{3}C{H}_3\left(g\right)$	$-84.67$

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

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=\left[\left(1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times ∆{\mathrm{H}}_{\mathrm{f},C{H}_{3}C{H}_3\left(g\right)}^0\right)\right]-\left[\left(1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times ∆{\mathrm{H}}_{\mathrm{f},HC\equiv CH \left(g\right)}^0\right)+\left(2\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times ∆{\mathrm{H}}_{\mathrm{f},H_2\left(g\right)}^0\right)\right]$

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=\left[\left(1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times -84.67\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)\right]-\left[\left(1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times \left(226.7\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)\right)+\left(2\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times 0.0\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)\right]$

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=\left(-84.67\right)-\left(226.7\right)$

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=-311.37 kJ$

Equation A: $HC\equiv CH \left(g\right)+2H_2\left(g\right)\to C{H}_{3}C{H}_3\left(g\right), ∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=-311.37 kJ$

Calculation for $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}$ for equation B:

$H_{2}C=C{H}_2 \left(g\right)+H_2\left(g\right)\to C{H}_{3}C{H}_3\left(g\right)$

	$∆{\mathrm{H}}_{\mathrm{f}}^0(\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l})$
$H_{2}C=C{H}_2 \left(g\right)$	$52.4$
$H_2\left(g\right)$	$0.0$
$C{H}_{3}C{H}_3\left(g\right)$	$-84.67$

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

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=\left[\left(1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times ∆{\mathrm{H}}_{\mathrm{f},C{H}_{3}C{H}_3\left(g\right)}^0\right)\right]-\left[\left(1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times ∆{\mathrm{H}}_{\mathrm{f},H_{2}C=C{H}_2 \left(g\right)}^0\right)+\left(1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times ∆{\mathrm{H}}_{\mathrm{f},H_2\left(g\right)}^0\right)\right]$

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=\left[\left(1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times -84.67\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)\right]-\left[\left(1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times \left(52.4\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)\right)+\left(1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\times 0.0\frac{\mathrm{k}\mathrm{J}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)\right]$

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=\left(-84.67\right)-\left(52.4\right)$

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=-137.07 kJ$

Equation B: $H_{2}C=C{H}_2 \left(g\right)+H_2\left(g\right)\to C{H}_{3}C{H}_3\left(g\right), ∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=-137.07 kJ$

For the final reaction, the ethylene ($H_{2}C=C{H}_2$) is on the product side. Therefore, we reverse the equation B. Reversing equation B means we need to reverse the sign of $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}$ of equation B. When we add equations A and B, ethane ($C{H}_{3}C{H}_3$) gets cancelled out and we get equation C.

Equation A: $HC\equiv CH \left(g\right)+2H_2\left(g\right)\to C{H}_{3}C{H}_3\left(g\right), ∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=-311.37 kJ$

Equation B: $C{H}_{3}C{H}_3\left(g\right)\to H_{2}C=C{H}_2 \left(g\right)+H_2\left(g\right), ∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=137.07 kJ$

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Final reaction: $HC\equiv CH \left(g\right)+H_2\left(g\right)\to H_{2}C=C{H}_2 \left(g\right), ∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}=-174.30 kJ$

$∆H_{rxn}$ for the hydrogenation acetylene with one mole of hydrogen gas to make ethylene is $-174.30 kJ$

Conclusion:

Applying $∆H_f^0$ values from Appendix 4 and Hess’s law, the $∆H_{rxn}$ for hydrogenation of acetylene ($C_2{H}_2$) with one mole of hydrogen gas to make ethylene ($C_2{H}_4$) is calculated.