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# .16 g of methane was subjected to cmbustion at 27 C in a bomb calorimeter.The temp. of calorimeter system(including water was found to rise by .5 C. Calculate the heat of combustion of methane at1 constant pressureConstant VolumeThe thermal capcity of calorimeter system is 17.7 J/K (R=25/3 J/mol k

Asked by Akhandpratapsingh1999 2nd January 2017, 8:21 PM

Heat of combustion at constant volume,

ΔE = Heat capacity of calorimeter system× rise in temperature × Mol.mass of compound / mass of compound

= 17.7 × 0.5 × 16/0.16 = 885

ΔE = -885 kJ mol-1

CH4(g) + 2O2(g) ——> CO2 (g) + 2H2O(l)

Δn = 1 – 3 = -2, T = 300 K, R = 8.314 × 10-3 kJ K-1mol-1

ΔH = ΔE + ΔnRT= –885 + (–2) × 8.314 × 10–3 kJk–1 mol–1

ΔH = ΔE + ΔnRT  = –885 + (–2) × 8.314 × 10–3 × 300

Answered by Expert 3rd January 2017, 10:58 AM
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