Question

the heat capacity of a bomb calorimeter is found to be 5.68 kj/°c . when a 1.652 g sample of glucose, c6h12o6 , is burned in the calorimeter, the temperature of the calorimeter increases from 24.56 °c to 29.10 °c. calculate the ∆h for the following reaction: c6h12o6(s) 6o2(g) → 6co2(g) 6h2o(g)

Answer 1

The ∆ H of combustion reaction of C6H12O6(s) + 6O2(g) => 6CO2(g) +6H2O(g) is -42.6 kJ

Given

Heat capacity of glucose= 5.68 kj/°c

The mass of glucose = 1.652 g

Temperature changes  =  24.56°C to 29.10°C

q = C mass ∆T

q = ∆ H

q = (5.68)(1.652)(29.1-24.56)

q = -42.6 kj

The heat energy released when one mole of a substance burns entirely in oxygen is known as the enthalpy of combustion of a substance. The temperature rise is measured as fuel burns. Calculating the enthalpy change of the reaction can be done using the mass of fuel that corresponds to the rise in temperature. Because combustion processes are exothermic, the enthalpy change (∆ H) value is always negative.

Hence, combustion reaction is exothermic reaction.

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