Question

The combustion of 0.590 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the temperature of the calorimeter by 2.125naughtC. The chamber was then emptied and recharged with 1.400 g of glucose (MW = 180.16 g/mol) and excess oxygen. How much did the temperature change from the combustion of the glucose? ΔHcomb for glucose is 2,780 kJ/mol.

Answer 1

2.943 °C temperature change from the combustion of the glucose has been taken place.

Step-by-step explanation:

Heat released on combustion of Benzoic acid; :

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.590 g

Moles of benzoic acid =
`(0.590 g)/(122.12 g/mol)=0.004831 mol`

Energy released by 0.004831 moles of benzoic acid on combustion:

`Q=3,228 kJ/mol * 0.004831 mol=15.5955 kJ=15,595.5 J`

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.125°C

`Q=C* \Delta T`

`15,595.5 J=C* 2.125^oC`

`C=7,339.05 J/^oC`

Heat released on combustion of Glucose: :

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=1.400 g

Moles of glucose =
`(1.400 g)/(180.16 g/mol)=0.007771 mol`

Energy released by the 0.007771 moles of calorimeter combustion:

`Q'=2,780 kJ/mol * 0.007771 mol=21.6030 kJ=21,603.01 J`

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

`Q'=C* \Delta T'`

`21,603.01 J=7,339.05 J/^oC* \Delta T'`

`\Delta T'=2.943^oC`

2.943 °C temperature change from the combustion of the glucose has been taken place.