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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 at
1 constant pressure
Constant Volume
The 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
Answered by Expert
Answer:

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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