Fructose, C_6 H_12 O_6 [s], is a sugar found in fruits and a source of energy fo
ID: 508502 • Letter: F
Question
Fructose, C_6 H_12 O_6 [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: C_6 H_12 O_s (s) + 6 O (g) rightarrow 6 CO_2 (g) + 6 H_2 O (I) 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 Delta H_rxn = Delta E_rxn. The Apollo Lunar Module was powered by a reaction similar to the following: 2 N_2 H_4 (I) + N_2 O_4 (I) rightarrow 3 N_2 (g) + 4 H_2 O (g) Using the values of Delta H degree given in Appendix B of your text, calculate the value of Delta H_rxn degree for this reaction. Instant cold packs used to ice athletic injuries on the field contain ammonium nitrate and water separated by a thin plastic divider. When the divider is broken, the ammonium nitrate dissolves according to the endothermic reaction: NH_4 NO_ (s) rightarrow NH_4^+ (aq) + NO_3^- (aq) In order to measure the enthalpy change for this reaction, 1.25 g of NH_4 NO_3 is dissolved in enough water to make 25.0 mL of solution. The initial temperature is 25.8 degree C and the final temperature (after the solid dissolves) is 21.9 degree C. Calculate the change in enthalpy for the reaction in kJ.Explanation / Answer
(1)
Heat change = Heat capacity of calorimeter * cahnge in temperature
q = 29.7 * 2.635
q = 78.26 kJ
This is the amount of heat released by 5.00 g. of fructose.
then, amount of heat released by 1 g. of fructose = 78.26 / 5.00 = 15.6 kJ
Then, amount of heat released by 180 g. (or) 1 mol of fructose = 15.6 * 180 = 2808 kJ
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