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1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation C_("(graphite)") + O_(2(g)) to CO_(2(g)) During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7 kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm ?

Answer» <html><body><p></p>Solution :Suppose q is the quantity of heat from the reaction <a href="https://interviewquestions.tuteehub.com/tag/mixture-1098735" style="font-weight:bold;" target="_blank" title="Click to know more about MIXTURE">MIXTURE</a> and `C_(V)` is the heat capacity of the calorimeter, then the quantity of heat absorbed by the calorimeter : <br/> `q=C_(v) xx Delta T` <br/> Quantity of heat from the reaction will have the same magnitude but opposite <a href="https://interviewquestions.tuteehub.com/tag/sign-1207134" style="font-weight:bold;" target="_blank" title="Click to know more about SIGN">SIGN</a> because the heat lost by the system is equal to the heat gained by the calorimeter. <br/> `q=-C_(v) xx Delta T` <br/> `=-20.7 "kJ/K" K xx (299- 298) K = -20.7 "kJ"` <br/> Here, negative sign indicates the exothermic nature of the reaction. <br/> <a href="https://interviewquestions.tuteehub.com/tag/thus-2307358" style="font-weight:bold;" target="_blank" title="Click to know more about THUS">THUS</a>, `Delta U` for the combustion of the 1g of graphite `= -20.7 "kJ K"^(-1)` <br/> For combustion of 1 <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> of graphite, <br/> `= (12.0 g "mol"^(-1) xx (-20.7 "kJ") )/(1g)` <br/> `= -2.48 xx 10^(2) "kJ mol"^(-1)` since `Deltan_(g) =0` <br/> `therefore Delta H = Delta U = - 2.48 xx 10^(2) "kJ mol"^(-1)`</body></html>


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