Chemistry - Thermochemistry Question with Solution | TestHub
A volume of 1.642 l sample of a mixture of methane gas and oxygen measured at 298 K and 1.192 atm, was allowed to react at constant pressure in a calorimeter which together with its content had a heat capacity of 1260 cal/K. The complete combustion of methane to carbon dioxide and water caused a temperature rise in calorimeter 0.667 K. The volume per cent of methane in original mixture is (Given the heat of combustion of methane is -210 kcal/mole)
Answer:
Solution:
The total number of moles, n, can be calculated using the ideal gas law PV=nRT:
n = PV/RT = (1.192 × 1.642)/(0.0821 × 298) = 0.0800 mol
The heat released is 1260 × 0.667 = 840.4 cal, which is equivalent to 0.840 kcal.
Therefore, the moles of CH₄ can be calculated as:
nCH₄ = 0.840/210 = 0.004 mol
The volume percentage, which is equal to the mole percentage, is:
Volume per cent = 0.004/0.0800 × 100 = 5%
The final answer is 5.
