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The heat of combusion of benzene in a bo...

The heat of combusion of benzene in a bomb calorimeter (i.e constant volume) was found to be `3263.9kJ mo1^(-1)` at `25^(@)C` Calculate the heat of combustion of benzene at constan pressure .

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The reaction is `C_(6)H_(6)(l) + 71/2 O_(2)g to 6CO_(2)(g) + 3H_2O + 3h_2O (l)`
In this reaction, `O_2` is the only gaseous reactant and `CO_(2)` is the only gaseous product.
`:. " " Delta n_(g) = n_p - n_r = 6 - 7 1/2 = -1 1/2 = -3/2`
Also, we are given `Delta U` (or `q_(v)) = 3263.9 kJ mol^(-1)`
`T = 25^@C = 298K`
`R = .314 K(-1) mol^(-1) = (8.314)/(1000) kJ K^(-1) mol^(-1)`
`DeltaH ("or" q_p) = Delta U + Delta n_(g)RT = -3263.9 kJ mol^(-1) + (3/2 mol) ((8.314)/(1000) kJ K^(-1)mol^(-1)) (298K)`
`=-3263.9 - 3.7 kJ mol^(-1) = -3267.6 kJ mol^(-1)`.
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