1.250 g of a sample of octane (C_(8)H_(18)) is burnt in excess of oxygen in a bomb calorimeter. The temperature of the calorimeter rises from 298 K to 304.73 K. If heat capacity of calorimeter is 8.93 kJ/K. Calculate the heat transferred. Also calculate Delta U and Delta H of the reaction at 298 K. The reaction involved is C_(8)H_(18(l)) + (25)/(2)O_(2(g)) rarr 8CO_(2(g)) + 9H_(2)O_((l)).

1.250 g of a sample of octane (C_(8)H_(18)) is burnt in excess of oxyg

1.	1.250 g of a sample of octane (C_(8)H_(18)) is burnt in excess of oxygen in a bomb calorimeter. The temperature of the calorimeter rises from 298 K to 304.73 K. If heat capacity of calorimeter is 8.93 kJ/K. Calculate the heat transferred. Also calculate Delta U and Delta H of the reaction at 298 K. The reaction involved is C_(8)H_(18(l)) + (25)/(2)O_(2(g)) rarr 8CO_(2(g)) + 9H_(2)O_((l)).
Answer» Solution :Let q be the QUANTITY of heat transferred to the calorimeter. Here, `Delta T = 304.73 - 298.08 = 6.73 K` Heat absorbed by the calorimeter `= C_(v) xx Delta T` `= 8.93(KJ K^(-1)) xx 6.73 (K) = 60.1 kJ` Molar mass of octane `(C_(8)H_(18))` Heat produced by 1.250 g of `C_(8)H_(18) = 60.1 kJ` Heat produced by 1 mol `= 114g` of `C_(8)H_(18)` `= (60.1 xx 114)/(1.250) = 5481.1 kJ` Thus, `Delta U = -5481.1 kJ mol^(-1)` `Delta n_((g))` of the reaction `= 8 - (25)/(2) = -4.5` `Delta H = Delta U + Delta nRT` `= -5481 + (-4.5) xx 8.314 xx 10^(-3) xx 298` `= -5492.2 kJ mol^(-1)`