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_{2}H_{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_{2}O(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 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 |
The balanced equation is
2 C_{2}H_{2}(g) + 5 O_{2}(g) → 4 CO_{2}(g) + 2 H_{2}O(g)The necessary heats of formation are
ΔH°_{f} [C_{2}H_{2}(g)] = 227.4 kJ/mol ΔH°_{f} [H_{2}O(g)] = -241.8 kJ/mol
ΔH°_{f} [CO_{2}(g)] = -393.5 kJ/mol
The standard enthalpy change for the reaction is
ΔH° = [4 ΔH°_f (CO_2)] + 2 ΔH°_f (H_2O)] – [2 ΔH°_f (C_2H_2)]
= (4 mol)(-393.5 kJ/mol) + (2 mol)(-241.8 kJ/mol) – (2 mol)(227.4 kJ/mol)
= -2512.4 kJ
