Chapter 2, Problem 42P

Given ΔH° for the reaction

${\text{H}}_2\text{(g)+}\frac{1}{2}{\text{O}}_2\text{(g)}\to {\text{ H}}_2\text{O(l)}$ $\text{ΔH° = -286 kJ}$ along with the information that the heat of combustion of ethane is $1560$ kJ/mol and that of ethylene is $1410$ kJ/mol, calculate $\text{ΔH°}$ for the hydrogenation of ethylene:

${\text{H}}_2{\text{C=CH}}_2{\text{(g)+H}}_2\text{(g)}\to {\text{CH}}_3{\text{CH}}_3\text{(g)}$

If the heat of combustion of acetylene is $1300$ kJ/mol, what is the value of $\text{ΔH°}$ for its hydrogenation to ethylene? To ethane?

What is the value of $\text{ΔH°}$ for the hypothetical reaction

${\text{2H}}_2{\text{C=CH}}_2\text{(g)}\to {\text{CH}}_3{\text{CH}}_3\text{(g)+HC}\equiv \text{CH(g) }$

Interpretation Introduction

Interpretation:

The heat of hydrogenation of ethylene and of acetylene to ethylene and ethane is to be calculated, and the enthalpy of the given hypothetical reaction is to be calculated.

Concept introduction:

Heat of a reaction can be determined from the measured heats of other reactions.

When a reaction equation is multiplied by some number, each coefficient in the balanced equation along with the associated enthalpy change is multiplied by that number.

If a reaction is reversed, the sign of the associated heat of reaction changes.

Answer to Problem 42P

Solution:

The heat of hydrogenation of ethylene is $\text{-136}$ kJ/mol.

${\text{ΔH}}^{\text{o}}$ for hydrogenation of acetylene to ethylene is $\text{-176}$ kJ/mol.

${\text{ΔH}}^{\text{o}}$ for hydrogenation of acetylene to ethane is $\text{-312}$ kJ/mol.

${\text{ΔH}}^{\text{o}}$ for the hypothetical reaction ${\text{2H}}_{\text{2}}{\text{C=CH}}_{\text{2}}\text{(g) }\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{(g) + C}}_{\text{2}}{\text{H}}_{\text{2}}\text{(g) }$ is $+40\text{ kJ}$.

Explanation of Solution

The combustion reactions of hydrogen (given), ethane, and ethylene are as follows:

$\begin{array}{l}{\text{H}}_{\text{2}}\text{(g) + }\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}\to {\text{H}}_{\text{2}}\text{O (l)}\Delta {\text{H}}^{\text{o}}\text{= }-\text{286 kJ/mol }\mathrm{...}\text{ (1)}\\ {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{(g) + }\frac{7}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}\to {\text{2CO}}_{\text{2}}\text{(g) + 3H}\text{O}\text{(l)}\text{ΔH=}-\text{1560kJ/mol }\mathrm{...}\text{ (2)}\\ {\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{(g) + 3O}}_{\text{2}}\text{(g)}\to {\text{2CO}}_{\text{2}}{\text{(g) + 2H}}_{\text{2}}\text{O(l)}\text{ΔH=}-\text{1410kJ/mol }\mathrm{...}\text{ (3)}\end{array}$

The reaction for the heat of hydrogenation of ethylene is shown below:

${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{(g) + H}}_{\text{2}}\text{(g) }\to {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{(g)}$

Using the above three equations, the heat of reaction for hydrogenation of ethylene is calculated as below:

Add reactions (1) and (3)m and add a reverse reaction (2):

$\begin{array}{l}{\text{ H}}_{\text{2}}\text{(g) + }\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}\to {\text{H}}_{\text{2}}\text{O(l)}\Delta {\text{H}}^{\text{o}}\text{= }-\text{286 kJ/mol}\\ {\text{+ C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{(g) + 3O}}_{\text{2}}\text{(g)}\to {\text{2CO}}_{\text{2}}{\text{(g) + 2H}}_{\text{2}}\text{O(l)}\text{ ΔH =}-\text{1410kJ/mol}\\ {\text{+ 2CO}}_{\text{2}}\text{(g) + 3H}\text{O}\text{(l) }\to {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{(g) + }\frac{7}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}\text{ΔH =}-\text{1560kJ/mol}\end{array}$

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${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{(g)+H}}_{\text{2}}\text{(g)}\to \text{C}\text{H}{}_{\text{6}}\text{(g) }\Delta \text{H = }-\text{136 kJ}$

(Note: Subtracting reaction (2) is the same as reversing and adding it. The final equation is obtained by cancelling out any terms common to both sides of the equation.)

Thus, the heat of hydrogenation of ethylene is $\text{-136}$ kJ/mol

b) The equation for heat of combustion of acetylene is as follows:

${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\text{(g) + 2}\frac{1}{2}{\text{O}}_{\text{2}}\text{(g) }\to {\text{2CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(l) }\Delta \text{H = }-\text{ 1300 kJ/mol }\mathrm{...}\text{ (4)}$

The reaction for heat of hydrogenation of acetylene to ethylene is shown below:

${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{(g) }\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{4}}\text{(g)}$

${\text{ΔH}}^{\text{o}}$ for hydrogenation of acetylene to ethylene is calculated using this reaction and reactions (1) and (3) above. Reactions (4) and (1) are added, and reaction (3) is subtracted (or reversed and added):

$\begin{array}{l}{\text{ C}}_{\text{2}}{\text{H}}_{\text{2}}\text{(g) + }\frac{5}{2}{\text{O}}_{\text{2}}\text{(g) }\to {\text{ 2CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(l) }\Delta \text{H = }-\text{ 1300 kJ}\\ +{\text{ H}}_{\text{2}}\text{(g) + }\frac{1}{2}{\text{ O}}_{\text{2}}\text{(g) }\to {\text{ H}}_{\text{2}}\text{O(l) }\Delta \text{H = }-\text{ 286 kJ}\\ +{\text{ 2CO}}_{\text{2}}{\text{(g) + 2H}}_{\text{2}}\text{O(l) }\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{(g) + 3O}}_2\text{(g) }\Delta \text{H = +1410 kJ}\end{array}$

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${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{(g) }\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{4}}\text{(g) }\Delta \text{H = -176 kJ }\mathrm{...}\text{(5)}$

Thus, the enthalpy change for hydrogenation of acetylene to ethylene is $\text{-176}$ kJ/mol.

The equation for heat of combustion of acetylene to is as follows:

${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\text{(g) + 2}\frac{1}{2}{\text{O}}_{\text{2}}\text{(g) }\to {\text{2CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(l) }\Delta \text{H = }-\text{ 1300 kJ/mol }\mathrm{...}\text{ (4)}$

The reaction for heat of hydrogenation of acetylene to ethane is shown below:

${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{(g) + 2H}}_{\text{2}}\text{(g) }\to {\text{ C}}_{\text{2}}{\text{H}}_6\text{(g)}$

${\text{ΔH}}^{\text{o}}$ for hydrogenation of acetylene to ethane is calculated using reaction (4) and reactions (1) and (2) above. Reaction(1) is multiplied by 2 and added to (4), and reaction (2) is subtracted (or reversed and added):

$\begin{array}{l}{\text{ C}}_{\text{2}}{\text{H}}_{\text{2}}\text{(g) + }\frac{5}{2}{\text{O}}_{\text{2}}\text{(g) }\to {\text{ 2CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(l) }\Delta \text{H = }-\text{ 1300 kJ}\\ +{\text{ 2H}}_{\text{2}}{\text{(g) + O}}_{\text{2}}\text{(g) }\to {\text{ 2H}}_{\text{2}}\text{O(l) }\Delta \text{H = }-\text{ 572 kJ}\\ {\text{+ 2CO}}_{\text{2}}{\text{(g) + 3H}}_{\text{2}}\text{O(l) }\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{6}}\text{(g) + }\frac{7}{2}{\text{O}}_{\text{2}}\text{(g) }\Delta \text{H = }-\text{ 1560 kJ}\end{array}$

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${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{(g) + 2H}}_{\text{2}}\text{(g) }\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{6}}\text{(g) }\Delta \text{H = }-\text{ 312 kJ }\mathrm{.....}\text{ (6)}$

Thus, the enthalpy change for hydrogenation of acetylene to ethane is $\text{-312}$ kJ/mol.

c) ${\text{ΔH}}^{\text{o}}$ for the hypothetical reaction ${\text{2H}}_{\text{2}}{\text{C=CH}}_{\text{2}}\text{(g) }\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{(g) + C}}_{\text{2}}{\text{H}}_{\text{2}}\text{(g) }$ is calculated using reactions (5) and (6). Reaction (5) is multiplied by 2 and reversed, and reaction (6) is added to it:

$\begin{array}{l}\text{ 2C}\text{H}{}_{\text{4}}\text{(g) }\to {\text{ 2C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{(g) + 2H}}_{\text{2}}\text{(g) }\Delta \text{H = +352 kJ}\\ {\text{+ C}}_{\text{2}}{\text{H}}_{\text{2}}{\text{(g) + 2H}}_{\text{2}}\text{(g) }\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{6}}\text{(g) }\Delta \text{H= }-\text{312 kJ}\end{array}$

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${\text{2C}}_{\text{2}}{\text{H}}_{\text{4}}\text{(g) }\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{(g) + C}}_{\text{2}}{\text{H}}_{\text{2}}\text{(g) }\Delta \text{H = + 40 kJ}$