Question 14.15: A 1.000-gram sample of octane, C8H18(l), is burned in a calo......

A 1.000-gram sample of octane, C_{8}H_{18}(l), is burned in a calorimeter like that shown in Figure 14.18, and the observed temperature increase is 1.679 K. The total heat capacity of the calorimeter is c_{V,cal}= 28.46\ kJ·K^{–1} (where we write c_{V,cal} to emphasize that the value of the heat capacity of the calorimeter is that at constant volume in this case).
(a) Calculate the energy of combustion per gram and per mole of C_{8}H_{18}(l).
(b) Using the molar enthalpy of formation values listed in Table 14.3, show that ΔH_{rxn} ≈ ΔU_{rxn} for this reaction at 25°C.

TABLE 14.3 standard enthalpies of formation, ΔH^{\circ}_{f}, for various substances a  25°C
Substance Formula ΔH^{\circ}_{f}/kJ\cdot mol^{-1} Substance Formula ΔH^{\circ}_{f}/kJ\cdot mol^{-1}
aluminum oxide Al_{2}O_{3}(s) -1675.7 hydrogen fluoride HF(g) -273.3
ammonia NH_{3}(g) -45.9 hydrogen iodide HI(g) +26.5
benzene C_{6}H_{6}(l) +49.1 hydrogen peroxide H_{2}O_{2}(l) -187.8
benzoic acid C_{6}H_{5}COOH(s) -385.2 iodine vapor I_{2}(g) +62.4
bromine vapor Br_{2}(g) +30.9 magnesium carbonate MgCO_{3}(s) -1095.8
butane C_{4}H_{10}(g) -125.7 magnesium oxide MgO(s) -601.6
calcium carbonate CaCO_{3}(s) -1207.6 magnesium sulfide MgS(s) -346.0
carbon (diamond) C(s) +1.897 methane CH_{4}(g) -74.6
carbon (graphite) C(s) 0 methanol (methyl alcohol) CH_{3}OH(l)
CH_{3}OH(g)
-239.2
-201.0
carbon (buckminster fullerene) C_{60}(s) +2327.0 methyl chloride CH_{3}Cl(g) -81.9
carbon dioxide CO_{2}(g) -393.5 nitrogen dioxide NO_{2}(g) +33.2
carbon monoxide CO(g) -110.5 nitrogen oxide NO(g) +91.3
carbon tetrachloride CCl_{4}(l)
CCl_{4}(g)
-128.2
-95.7
dinitrogen tetroxide N_{2}O_{4}(g)
N_{2}O_{4}(l)
+11.1
-19.5
chromium (III) oxide Cr_{2}O_{3}(s) -1139.7 octane C_{8}H_{18}(l) -250.1
cyclohexane C_{6}H_{12}(l) -156.4 pentane C_{5}H_{12}(l) -173.5
ethane C_{2}H_{6}(g) -84.0 propane C_{3}H_{8}(g) -103.8
ethanol (ethyl alcohol) CH_{3}CH_{2}OH(l) -277.6 sodium carbonate Na_{2}CO_{3}(s) -1130.7
ethene (ethylene) C_{2}H_{4}(g) +52.4 sodium oxide Na_{2}O(s) -414.2
ethyne (acetylene) C_{2}H_{2}(g) +227.4 sucrose C_{12}H_{22}O_{11}(s) -2226.1
freon-12 (dichloro difluoromethane) CF_{2}Cl_{2}(g) -477.4 sulfur dioxide SO_{2}(g) -296.8
glucose C_{6}H_{12}O_{6}(s) -1273.3 sulfur trioxide SO_{3}(g) -395.7
hexane C_{6}H_{14}(l) -198.7 tin(IV) oxide SnO_{2}(s) -577.6
hydrazine N_{2}H_{4}(l)
N_{2}H_{4}(g)
+50.6
+95.4
water H_{2}O(l)
H_{2}O(g)
-285.8
-241.8
hydrogen bromide HBr(g) -36.3
hydrogen chloride HCl(g) -92.3
Data from CRC Handbook of Chemistry and Physics, 86th Ed., Ed. David R. Lide, CRC Press, 2005–2006. (More thermodynamic data are given in Appendix D.)
figure 14.18
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(a) The equation for the combustion reaction is

2\ C_{8}H_{18}(l) + 25\ O_{2}(g) → 16\ CO_{2}(g) + 18\ H_{2}O(l)

The energy of combustion is given by

\Delta U = –c_{V,cal}\Delta T

which is analogous to Equation 14.36 under conditions of constant volume.
Using this equation, we find that the value of Δ
U for 1.000 grams of C_{8}H_{18}(l) is

\Delta H = –c_{P,cal}\Delta T            (14.36)

\Delta U_{rxn}\ [per\ 1.000\ g\ C_{8}H_{18}(l)] = –(28.46\ kJ·K^{–1})(1.679\ K·g^{–1})= –47.78\ kJ·g^{–1}

because the observed ΔT is associated with the combustion of 1.000 grams of octane. The value of \Delta U_{rxn} per mole of octane (molar\ mass = 114.22\ g·mol^{–1}) is

\Delta U_{rxn}\ [per\ mol\ C_{8}H_{18}(l)] = (-47.78\ kJ·g^{–1})(114.22\ g·mol^{–1})= –5457\ kJ·mol^{–1}

or in terms of the overall reaction equation we write,

C_{8}H_{18}(l) + \frac{25}{2} O_{2}(g) → 8\ CO_{2}(g) + 9\ H_{2}O(l) \quad \Delta U_{rxn} = –5457\ kJ·mol^{–1}

Here we balanced the chemical equation using fractional coefficients rather than whole number coefficients because we are interested in the standard enthalpy of combustion per mole of octane.

(b) Using the values listed in Table 14.3 and applying Equation 14.24, we find that

ΔH^{\circ}_{rxn} =ΔH^{\circ}_{f}\ [products] − ΔH^{\circ}_{f}\ [reactants]            (14.24)

\Delta H_{rxn} = \{8\ ΔH^{\circ}_{f} [CO_{2}(g)] + 9\ ΔH^{\circ}_{f} [H_{2}O(l)]\} – \{ΔH^{\circ}_{f} [C_{8}H_{18}(l)] + \frac{25}{2} ΔH^{\circ}_{f} [O_{2}(g)]\}\\= \{8(–393.5\ kJ·mol^{–1}) + 9(–285.8\ kJ·mol^{–1})\} – \{(–250.1\ kJ·mol^{–1})\}\\= –5470.1\ kJ·mol^{–1}

Hence, \Delta H_{rxn} and \Delta U_{rxn} differ by two-tenths of a percent, showing that ΔH_{rxn} ≈ ΔU_{rxn} is a good approximation in this case.

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