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Question 8.9: Using Standard Heats of Formation to Calculate ΔH° Oxyacetyl......

Using Standard Heats of Formation to Calculate ΔH°

Oxyacetylene welding torches burn acetylene gas, C_2H_2(g). Use the information in Table 8.2 to calculate ΔH° in kilojoules for the combustion reaction of acetylene to yield CO_2(g) and H_2O(g).

STRATEGY

Write the balanced equation, look up the appropriate heats of formation for each reactant and product in Table 8.2, and then carry out the calculation, making sure to multiply each ΔH°_{f} by the coefficient given in the balanced equation. Remember also that ΔH°_{f}(O_2) = 0 kJ/mol.

Table 8.2 Standard Heats of Formation for Some Common Substances at 25°C

Substance Formula \Delta H^{\circ}_{f} (kJ/mol) Substance
Acetylene C_2H_2(g) 227.4 Hydrogen chloride
Ammonia NH_3(g) -46.1 Iron(III) oxide
Carbon dioxide CO_2(g) -393.5 Magnesium carbonate
Carbon monoxide CO(g) -110.5 Methane
Ethanol C_2H_5OH(l) -277.7 Nitric oxide
Ethylene C_2H_4(g) 52.3 Water (g)
Glucose C_6H_{12}O_6(s) -1273.3 Water (l)
fig8.9
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The balanced equation is

2 \mathrm{C}_2 \mathrm{H}_2(g)+5 \mathrm{O}_2(g) \longrightarrow 4 \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(g)

The necessary heats of formation from Table 8.2 are

\begin{aligned}& \Delta H^{\circ}{ }_{\mathrm{f}}\left[\mathrm{C}_2 \mathrm{H}_2(g)\right]=+227.4 \mathrm{~kJ} / \mathrm{mol} \quad \Delta H^{\circ}{ }_{\mathrm{f}}\left[\mathrm{H}_2 \mathrm{O}(g)\right]=-241.8 \mathrm{~kJ} / \mathrm{mol} \\& \Delta H^{\circ}{ }_{\mathrm{f}}\left[\mathrm{CO}_2(g)\right]=-393.5 \mathrm{~kJ} / \mathrm{mol}\end{aligned}

The standard enthalpy change for the reaction is

\begin{aligned}\Delta H^{\circ} & =\left[4 \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{CO}_2\right)+2 \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)\right]-\left[2 \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{C}_2 \mathrm{H}_2\right)\right] \\& =(4 \mathrm{~mol})(-393.5 \mathrm{~kJ} / \mathrm{mol})+(2 \mathrm{~mol})(-241.8 \mathrm{~kJ} / \mathrm{mol})-(2 \mathrm{~mol})(227.4 \mathrm{~kJ} / \mathrm{mol}) \\& =-2512.4 \mathrm{~kJ}\end{aligned}

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