2.00 grams of Mn is burned in a bomb calorimeter. The water\'s temperature rose

Question

2.00 grams of Mn is burned in a bomb calorimeter. The water's temperature rose from 20.00 degrees centigrade to 26.73 degrees C.   The heat capacity of the calorimeter is 2.500 kJ/oC. What is the enthalpy change (?H) for this reaction?
3Mn(s) + 2O2(g) -> Mn3O4(s)   ?H rxn = ?

-43.8 kJ/molrxn

-1387kJ/molrxn

-2.36 x 104kJ/molrxn

-7.01 x 102 kJ/molrxn

-20.9 kJ/molrxn

-602 kJ/molrxn

-1460 kJ/molrxn

-1.71 x 103 kJ/molrxn

-43.8 kJ/molrxn

-1387kJ/molrxn

-2.36 x 104kJ/molrxn

-7.01 x 102 kJ/molrxn

-20.9 kJ/molrxn

-602 kJ/molrxn

-1460 kJ/molrxn

-1.71 x 103 kJ/molrxn

Explanation / Answer

2.00 grams of Mn is burned in a bomb calorimeter. The water's temperature rose from 20.00 degrees centigrade to 26.73 degrees C.   The heat capacity of the calorimeter is 2.500 kJ/oC. What is the enthalpy change (?H) for this reaction?

heat added = heat capacity of calorimeter* change in temperature = 2.5*(26.73-20) KJ=16.825 KJ

since there is temperature rise, the reaction is exothemirc and deltaH= -16.825 Kj

moles of Mn= mass/atomic weight =2/55=0.036 mole

enthalpy change/mole =-16.825/0.036=-463 Kj/mole

since there is three moles of Mn, enthalpy change = -463*3= -1388 Kj ( second one is correct)

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