The heat liberated on complete combustion of 7.8g of benzene is 327kj, this heat has been measured at constant vol. and at 27 degree C. Calculate the heat of combustion of benzene at const. temp.
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48
We have, C6H6(l) + 15/2O2(g) → 6CO2(g) + 3H2O(l)
Therefore, ∆n = 6 – 15/2 = – 3/2
Also, ∆Uper mol = – (327×78)/7.8
= – 3270 kJ
Therefore, ∆H = ∆U + ∆nRT
= – 3270 + (– 3/2)×300×10–3
Or, ∆H = – 3273.735 (C)
Therefore, ∆n = 6 – 15/2 = – 3/2
Also, ∆Uper mol = – (327×78)/7.8
= – 3270 kJ
Therefore, ∆H = ∆U + ∆nRT
= – 3270 + (– 3/2)×300×10–3
Or, ∆H = – 3273.735 (C)
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30
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