A calorimeter contains 25.0 mL of water at 15.0 degree C. When 1.80 g of X(a sub
ID: 517773 • Letter: A
Question
A calorimeter contains 25.0 mL of water at 15.0 degree C. When 1.80 g of X(a substance with a molar mass of 540 g/mol) is added, it dissolves via the reaction and the temperature of the solution increases to 25.5 degree C. Calculate the enthalpy change, Delta H, for this reaction per mole of X. Assume that the specific heat of the resulting solution is equal to that of water [4.18J/(g. degree C)] that density of water 1.00 g/mL, and that no heat is lost to the calorimeter itself not to the surroundings. Express the change in enthalpy in kilojoules per mole to three significant figures. Consider the reaction C_12H_22O_ (s) + 12O_2(g) rightarrow 12CO_2(g) + 11H_2O(I) in which 10.0 g of sucrose, C_12H_22O_11 was in a bomb calorimeter with a capacity of 7.50 KJ/degree C. The temperature increase inside the calorimeter was found to be 22.0 degree C. Calculate the change in energy, Delta E for this reaction per mole of Express the change in internal energy in kilojoules per mole to three significant figures.Explanation / Answer
PART (A)
Given that,
Volume of water = 25.0 mL
density of the water = 1.0 g/mL
So, mass of water = 25.0 g.
Specific heat of water = 4.18 J/g.0C
Change in temperature = t2-t1 = 25.5 - 15.0 = 10.5 0C
therefore, heat change of water = m s (t2-t1)
q = 25.0 * 4.18 * 10.5
q = 1097.25 J
This amount og heat has been released from 1.50 g. of X
moles of X = mass / molar mass = 1.50 / 54.0 = 0.0278 mol
Enthalpy change per mole = heat change / number of moles
deltaH = q / n
deltaH = 1097.25 / 0.0278
deltaH = 39469 J/mol
delta H = 39.5 kJ/mol
PART(B)
change in heat = Ccal*deltaT = 7.50 * 22.0
q = 0.341 kJ
Moles of sucrose = mass / molar mass = 10.0 / 342. = 0.0292 mol
Change in internal energy = q / n
deltaE = 0.341 / 0.0292
deltaE = 11.7 kJ/mol
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