A volume of 2.463 l sample of a mixture of methane gas and oxygen measured at 300 K and 1.00 atm, was allowed to react at constant pressure in a calorimeter which together with its content had a heat

Question

A volume of 2.463 l sample of a mixture of methane gas and oxygen measured at 300 K and 1.00 atm, was allowed to react at constant pressure in a calorimeter which together with its content had a heat capacity of 1500 cal/K. The complete combustion of methane to carbon dioxide and water caused a temperature rise in calorimeter 0.40 K. The volume per cent of methane in original mixture is (Given the heat of combustion of methane is -200 kcal/mole)

Answer:

3.00

Solution:

Total moles n = PV/RT = 1.00 × 2.463/0.0821 × 300 = 0.100 mol. Heat released = 1500 × 0.40 = 600 cal = 0.60 kcal, so n_CH4 = 0.60/200 = 0.003 mol. Volume per cent = mole per cent = 0.003/0.100 × 100 = 3. Answer: 3.

TestHub · Accepted Answer

Total moles n = PV/RT = 1.00 × 2.463/0.0821 × 300 = 0.100 mol. Heat released = 1500 × 0.40 = 600 cal = 0.60 kcal, so n CH4 = 0.60/200 = 0.003 mol. Volume per cent = mole per cent = 0.003/0.100 × 100 = 3. Answer: 3.

Chemistry - Thermochemistry Question with Solution | TestHub

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