PART 1. A calorimeter contains 30.0 mL of water at 11.5 degrees C. When 2.40 g o
ID: 713526 • Letter: P
Question
PART 1. A calorimeter contains 30.0 mL of water at 11.5 degrees C. When 2.40 g of X (a substance with a molar mass of 62.0 g/mol) is added, it dissolves via the reaction X(s)+H2O(l)X(aq) and the temperature of the solution increases to 30.0 degrees C . Calculate the enthalpy change, H, for this reaction per mole of X. Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(g*degrees C)], that density of water is 1.00 g/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings. Express the change in enthalpy in kilojoules per mole to three significant figures.
PART 2. Consider the reaction: C12H22O11(s)+12O2(g)12CO2(g)+11H2O(l) in which 10.0 g of sucrose, C12H22O11, was burned in a bomb calorimeter with a heat capacity of 7.50 kJ/degrees C. The temperature increase inside the calorimeter was found to be 22.0 degrees C. Calculate the change in internal energy, E, for this reaction per mole of sucrose. Express the change in internal energy in kilojoules per mole to three significant figures.
Explanation / Answer
1.
Mass of water = density * volume = 1.00 * 30.0 = 30.0 g.
Specific heat of water = 4.18 J/g-degree C
Change in temperature = 30.0 - 11.5 = 18.5 degree C
Heat change = mass * specific heat * change in temperature
q = 30.0 * 4.18 * 18.5
q = 2.32 * 103 J
Moles of solute = mass / molar mass
n = 2.40 / 62.0
n = 0.0387 mol
Change in enthalpy = q / n
deltaH = 2.32 * 103 / 0.0387
deltaH = 59933 J
deltaH = 59.9 kJ
2.
Heat change = Heat capacity of calorimeter * change in temperature
q = 7.50 * 22.0
q = 165. kJ
Moles of sucrose = mass / molar mass
n = 10.0 / 342.
n = 0.0292 mol
Change in internal eenrgy = 165 / 0.0292
deltaE = 5.64 * 103 kJ
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