USING STANDARD HEATS OF FORMATION TO CALCULATE ΔH° Oxyacetylene welding torches burn acetylene gas, C2H2(g). Use the information in Table 8.2 to calculate ΔH° in kilojoules for the combustion reaction of acetylene to yield CO2(g) and H2O(g). STRATEGY Write the balanced equation, look up the

Question

USING STANDARD HEATS OF FORMATION TO CALCULATE ΔH°

Oxyacetylene welding torches burn acetylene gas, [latex]C_{2}H_{2}(g)[/latex]. Use the information in Table 8.2 to calculate ΔH° in kilojoules for the combustion reaction of acetylene to yield [latex]CO_{2}(g) and H_{2}O(g)[/latex].

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 [latex]ΔH°_{f}[/latex] by the coefficient given in the balanced equation. Remember also that [latex]ΔH°_{f}(O_{2}) = 0 kJ/mol[/latex].

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

Substance Formula [latex]ΔH°_{f}[/latex](kJ/mol)	Substance Formula [latex]ΔH°_{f}[/latex](kJ/mol)
Acetylene [latex]C_{2}H_{2}(g)[/latex] 227.4	Hydrogen chloride HCl(g) -92.3
Ammonia [latex]NH_{3}(g)[/latex] -46.1	Iron(III) oxide [latex]Fe_{2}O_{3}(s)[/latex] -824.2
Carbon dioxide [latex]CO_{2}(g)[/latex] -393.5	Magnesium carbonate [latex]MgCO_{3}(s)[/latex] -1095.8
Carbon monoxide CO(g) -110.5	Methane [latex]CH_{4}(g)[/latex] -74.8
Ethanol [latex]C_{2}H_{5}OH(l)[/latex] -277.7	Nitric oxide NO(g) 91.3
Ethylene [latex]C_{2}H_{4}(g)[/latex] 52.3	Water(g) [latex]H_{2}O(g)[/latex] -241.8
Glucose [latex]C_{6}H_{12}O_{6}(s)[/latex] -1273.3	Water(l) [latex]H_{2}O(g)[/latex] -285.8

Accepted Answer

The balanced equation is

[latex]2 C_{2}H_{2}(g) + 5 O_{2}(g) → 4 CO_{2}(g) + 2 H_{2}O(g)[/latex]

The necessary heats of formation are

[latex]ΔH°_{f} [C_{2}H_{2}(g)] = 227.4 kJ/mol ΔH°_{f} [H_{2}O(g)][/latex] = -241.8 kJ/mol

[latex]ΔH°_{f} [CO_{2}(g)][/latex] = -393.5 kJ/mol

The standard enthalpy change for the reaction is
[latex]ΔH° = [4 ΔH°_f (CO_2)] + 2 ΔH°_f (H_2O)] - [2 ΔH°_f (C_2H_2)][/latex]

= (4 mol)(-393.5 kJ/mol) + (2 mol)(-241.8 kJ/mol) - (2 mol)(227.4 kJ/mol)

= -2512.4 kJ

Substance Formula $ΔH°_{f}$ (kJ/mol)	Substance Formula $ΔH°_{f}$ (kJ/mol)
Acetylene $C_{2}H_{2}(g)$ 227.4	Hydrogen chloride HCl(g) -92.3
Ammonia $NH_{3}(g)$ -46.1	Iron(III) oxide $Fe_{2}O_{3}(s)$ -824.2
Carbon dioxide $CO_{2}(g)$ -393.5	Magnesium carbonate $MgCO_{3}(s)$ -1095.8
Carbon monoxide CO(g) -110.5	Methane $CH_{4}(g)$ -74.8
Ethanol $C_{2}H_{5}OH(l)$ -277.7	Nitric oxide NO(g) 91.3
Ethylene $C_{2}H_{4}(g)$ 52.3	Water(g) $H_{2}O(g)$ -241.8
Glucose $C_{6}H_{12}O_{6}(s)$ -1273.3	Water(l) $H_{2}O(g)$ -285.8

Question 8.9: USING STANDARD HEATS OF FORMATION TO CALCULATE ΔH° Oxyacetyl...

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