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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» <html><body><p></p>Solution :Let q be the <a href="https://interviewquestions.tuteehub.com/tag/quantity-1174212" style="font-weight:bold;" target="_blank" title="Click to know more about QUANTITY">QUANTITY</a> of heat transferred to the calorimeter. <br/> Here, `Delta T = 304.73 - 298.08 = 6.73 K` <br/> Heat absorbed by the calorimeter `= C_(v) xx Delta T` <br/> `= <a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>.93(<a href="https://interviewquestions.tuteehub.com/tag/kj-1063034" style="font-weight:bold;" target="_blank" title="Click to know more about KJ">KJ</a> K^(-1)) xx 6.73 (K) = 60.1 kJ` <br/> Molar mass of octane `(C_(8)H_(18))` <br/> Heat produced by 1.250 g of `C_(8)H_(18) = 60.1 kJ` <br/> Heat produced by 1 mol `= 114g` of `C_(8)H_(18)` <br/> `= (60.1 xx <a href="https://interviewquestions.tuteehub.com/tag/114-268556" style="font-weight:bold;" target="_blank" title="Click to know more about 114">114</a>)/(1.250) = 5481.1 kJ` <br/> Thus, `Delta U = -5481.1 kJ mol^(-1)` <br/> `Delta n_((g))` of the reaction `= 8 - (25)/(2) = -4.5` <br/> `Delta H = Delta U + Delta nRT` <br/> `= -5481 + (-4.5) xx 8.314 xx 10^(-3) xx 298` <br/> `= -5492.2 kJ mol^(-1)`</body></html>


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