Chemistry - Thermochemistry Question with Solution | TestHub
When 0.1 mole of C6H14(l) at 300 K is completely burned at constant pressure in some oxygen gas at 300 K, yielding products gaseous H2O, CO and CO2 at 700 K, the process yielding 75.16 kcal of heat to the surrounding. Given:
Δf H (kcal/mol): C6H14(l) = -40.0, CO2(g) = -94.0, CO(g) = -26.0, H2O(g) = -58.0
Molar heat capacity at constant pressure (cal/K-mol): CO(g) = 7.0, CO2(g) = 9.0, H2O(g) = 8.0
Assume that all Δr H are independent of temperature and R = 2 cal/K-mol.
How many moles of CO2 are produced?
Options:
Answer:
Solution:
0.1 mol C6H14 gives 0.6 mol C atoms and 0.7 mol H2O. Let CO2 = a and CO = 0.6 - a. Step 1 at 300 K: ΔH1 = -94a - 26(0.6-a) - 58(0.7) + 0.1(40) = -68a - 15.6 - 40.6 + 4.0 = -68a - 52.2 kcal. Step 2, heat products by 400 K: ΔH2 = [9a + 7(0.6-a) + 8(0.7)](0.4) = (2a + 9.8)(0.4) = 0.8a + 3.92 kcal. Total = -67.2a - 48.28 = -75.16, so a = 26.88/67.2 = 0.4 mol (CO = 0.2 mol). This is option (c).
