1. Fructose, C6H12O6 (s), is a sugar found in fruits and a source of energy for

Question

1. Fructose, C6H12O6 (s), is a sugar found in fruits and a source of energy for the body. The combustion of fructose takes place according to the following equation:

C6H12O6 (s) + 6 O2 (g)a6 CO2 (g) + 6 H2O (l)

When 5.00 grams of fructose are burned in excess oxygen in a bomb calorimeter with a heat capacity of 29.7 kJ/K, the temperature of the calorimeter increases by 2.635 K. Calculate the standard enthalpy of combustion per gram and per mole of fructose. Assume that Hrxn Erxn. HINT: This problem is using Kelvin as the temperature unit – treat this problem as you would the ones we did in class, but use the Kelvin scale for temperature.

Explanation / Answer

heat given by fructose = heat absorbed by calorimeter

= heat capacity x rise in temperature

= 29.7 kJ/K x 2.635K

= 78.2595 kJ

5 g of fructose on combution gave 78.2595 kJ of heat

a) enthalpy of combustion per gram of fructose = 78.2595 /5

= 15. 6519 kJ/g

b) enthalpy of combustion per mole of fructose = 15.6519 kJ/g x 180g/mol

= 2817.342 kJ/mol

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