Chemistry - Thermochemistry Question with Solution | TestHub
A volume of 4.0 l of a mixture of ethylene and methane gases on complete combustion at 300 K produced 6.0 l of carbon dioxide. Find out the amount of heat evolved on burning the 4.0 l of gaseous mixture (at 1 atm and 300 K). The heats of combustion of ethylene and methane are -1573 and -890 kJ per mole, respectively, at 300 K.
Options:
Answer:
Solution:
Let the mixture contain x liters of ethylene and y liters of methane.
The total volume is 4 liters, so we have the equation:
x + y = 4
Each liter of ethylene produces 2 liters of CO₂ and each liter of methane produces 1 liter of CO₂. The total CO₂ produced is 6 liters, leading to the equation:
2x + y = 6
Solving these two equations, we find x = 2 liters and y = 2 liters.
The molar volume at 300 K and 1 atm can be calculated using the ideal gas law PV = nRT, where V/n = RT/P.
Vmolar = 0.0821 L atm mol-1 K-1 × 300 K = 24.63 L/mol
Thus, the number of moles for each gas is:
n = 2 L/(24.63 L/mol) ≈ 0.0812 mol
The heat released is the sum of the heats of combustion for both gases. Assuming the given values (1573 and 890) are molar heats of combustion in kJ/mol:
Heat = 0.0812 mol × (1573 kJ/mol + 890 kJ/mol)
Heat = 0.0812 mol × 2463 kJ/mol ≈ 200 kJ
This corresponds to option (d).
