Question

The combustion of 0.570 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 2.053naughtC. The chamber was then emptied and recharged with 2.900 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

The temperature change from the combustion of the glucose is 6.097°C.

Step-by-step explanation:

Benzoic acid;

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

Mass of benzoic acid = 0.570 g

Moles of benzoic acid =
`(0.570 g)/(122.12 g/mol)=0.004667 mol`

Energy released by 0.004667 moles of benzoic acid on combustion:

`Q=3,228 kJ/mol * 0.004667 mol=15.0668 kJ=15,066.8 J`

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

`Q=C* \Delta T`

`15,066.8 J=C* 2.053^oC`

`C=7,338.92 J/^oC`

Glucose:

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

Mass of glucose=2.900 g

Moles of glucose =
`(2.900 g)/(180.16 g/mol)=0.016097 mol`

Energy released by the 0.016097 moles of calorimeter combustion:

`Q'=2,780 kJ/mol * 0.016097 mol=44.7491 kJ=44,749.1 J`

Heat capacity of the calorimeter = C (calculated above)

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

`Q'=C* \Delta T'`

`44,749.1 J=7,338.92 J/^oC* \Delta T'`

`\Delta T'=6.097^oC`

The temperature change from the combustion of the glucose is 6.097°C.