Chemistry - Thermochemistry Question with Solution | TestHub
When 0.1 mole of C8H18(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 800 K, the process yielding 87.3 kcal of heat to the surrounding. Given:
Δf H (kcal/mol): C8H18(l) = -74.0, CO2(g) = -94.0, CO(g) = -26.5, H2O(g) = -58.0
Molar heat capacity at constant pressure (cal/K-mol): CO(g) = 7.0, CO2(g) = 8.0, H2O(g) = 6.0
Assume that all Δr H are independent of temperature.
How many moles of CO2 are produced?
Options:
Answer:
Solution:
0.1 mol C₈H₁₈ gives 0.8 mol C atoms and 0.9 mol H₂O. Let the moles of CO₂ be a and the moles of CO be 0.8 - a.
Step 1: Calculate the enthalpy change, Δ H1, for the reaction at 300 K.
Δ H1 = -94a - 26.5(0.8-a) - 58(0.9) - 0.1(-74)
Δ H1 = -94a - 21.2 + 26.5a - 52.2 + 7.4
Δ H1 = -67.5a - 66.0 kcal
Step 2: Calculate the enthalpy change, Δ H2, to heat the products from 300 K to 800 K (a temperature change of 500 K).
Δ H2 = [8a + 7(0.8-a) + 6(0.9)](0.5)
Δ H2 = [8a + 5.6 - 7a + 5.4](0.5)
Δ H2 = [a + 11](0.5)
Δ H2 = 0.5a + 5.5 kcal
The total enthalpy change, Δ H, is the sum of Δ H1 and Δ H2. We are given that 87.3 kcal is released, so Δ H = -87.3 kcal.
Δ H = Δ H1 + Δ H2
-87.3 = (-67.5a - 66.0) + (0.5a + 5.5)
-87.3 = -67.0a - 60.5
-87.3 + 60.5 = -67.0a
-26.8 = -67.0a
a = 26.8/67.0
a = 0.4 mol CO₂
Therefore, the moles of CO are also 0.8 - 0.4 = 0.4 mol. This corresponds to option (c).
