When 1.00 g of glucose, C_6H_12O_6, is combusted, 15.57 kJ of heat is released.

Question

When 1.00 g of glucose, C_6H_12O_6, is combusted, 15.57 kJ of heat is released. A 2.500-g sample of glucose is combusted in a bomb calorimeter, raising the temperature from 20.55 degree C to 23.25 degree C. What is the heat capacity of the calorimeter in kJ/K? If the same calorimeter is then used to combust a 1.250-g sample of octane, C_8H_18, and the temperature in the calorimeter rises 5.15 degree C, determine Delta_r U for the reaction as written: 2 C_8H_18(l) + 25 O_2(g) rightarrow 16 CO_2 + 18H_2O Determine Delta_r U for the reaction as written: 4 C_8H_18(l) + 50 O_2(g) rightarrow 32 CO_2 + 36 H_2O

Explanation / Answer

heat liberted when 2.5 gm of glucose is combusted = 2.5*15.57 Kj=38.925 kj

heat liberated = heat capacity of calorimter*temperature difference

38.925*1000 = heat capacity of calorimter*(23.25-20.55), heat capacity of calorimeter =14417 J/deg.c= 14.417 Kj/deg.c

enthalpy change = heat capacity of calorimeter* temperature difference = 14.418*5.25=75.69 Kj

moles of octane in 1.25 gm = mass/molar mass = 1.25/114=0.0109

enthalpy change/moles = 75.69/0.0109=6903 Kj/mole

enthalpy change= change in internal energy + deltan*R*deltaT

6903= change in internal energy+(36+32-50) *5.15*8.314

6903-770 = 6125 Kj/mole

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