Fructose C_6H_12O_6_(s), is a sugar found in fruits and a source of energy for t
ID: 509954 • Letter: F
Question
Fructose C_6H_12O_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_6H_12O_6(s) + 6 O_2 (g) rightarrow 6 CO_2 (g) + 6 H_2O (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 Hrxn = delta E_rxn The Apollo Lunar Module was powered by a reaction similar to the following: 2 N_2H_4 (l) + N_2O_4 (I) rightarrow 3 N_2 (g) + 4 H_2O (g) Using the values of delta H_f 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_4NO_3 (s) rightarow NH_4^+ (aq) + NO_3^- (aq) In order to measure the enthalpy change for this reaction, 1.25 g of NH_4NO_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
Heat released =Heat capacity of calorimeter * change in temperature
q = 29.7 * 2.635
q = 78.26 kJ
Amount of heat liberated per gram = 78.26 / 5 = 15.65 kJ/g.
Moles of fructose used = mass / molar mas = 5.0 / 180 = 0.0278 mol
Therefore, amount of heat liberated per mol of fructose =78.26/0.0278=2815.1 kJ/mol=2.82*103kJ/mol
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