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1 g of graphiteis burnt in a bombcalorimeter in excess of oxygen at 298 K and 1 atmosphericpressure accordingto the equation C ( graphite ) + O_(2)(g) rarr CO_(2)(g) During the reaction, temperature rises from298K to 299 K. If the heat capacity ofthe bomb calorimeter is 20.7 kJ // K , what is the enthalpy change for the above reaction at 298K and 1 atm ?

Answer» <html><body><p></p>Solution :Rise in <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> of the calorimeter `= 299- 298K = <a href="https://interviewquestions.tuteehub.com/tag/1k-282789" style="font-weight:bold;" target="_blank" title="Click to know more about 1K">1K</a>` <br/> Heat <a href="https://interviewquestions.tuteehub.com/tag/capacity-908575" style="font-weight:bold;" target="_blank" title="Click to know more about CAPACITY">CAPACITY</a> of the calorimeter `= 20.7 k J K^(-1)` <br/> `:. `Heat absorbed by the calorimeter `= C_(v) xxDelta T= (20.7 k J K^(-1)) (1K) = 20.7 kJ` <br/> This s the heat evolved in the combustionof1 g of graphite. <br/> `:. `Heat evolved inthe combustionof 1 <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> of graphite , `i.e., 12g` of graphite `= 20.7 xx 12 kJ = 248.4 kJ` <br/>As this is the heat evolved and the vessel is <a href="https://interviewquestions.tuteehub.com/tag/closed-919469" style="font-weight:bold;" target="_blank" title="Click to know more about CLOSED">CLOSED</a>, therefore, enthalpy change of the reaction `( Delta U )` <br/> `= - 248.4 kJ mol^(-1)`</body></html>


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