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.
What is the magnitude of work done by the system?
Options:
Answer:
Solution:
The gaseous reactant is O₂ (0.85 mol) at 300 K. Note that C₆H₁₄ is a liquid.
The gaseous products are 0.4 + 0.2 + 0.7 = 1.3 mol at 700 K.
The work done, |w|, can be calculated using the formula PΔ V = R(nf Tf - ni Ti).
|w| = R(nf Tf - ni Ti)
Substituting the given values:
|w| = 2(1.3 × 700 - 0.85 × 300)
|w| = 2(910 - 255)
|w| = 1310 cal
|w| = 1.31 kcal
This corresponds to option (b).
