A student conducts an experiment to determine the enthalpy of solution for lithium chloride dissolved in water. the student combines 5.00 grams of lithium chloride with 100.0 ml of distilled water. the initial temperature of the water is 23.0°c and the highest temperature after mixing reaches 33.0°c. assume a density of 1.00 g/ml and a specific heat of 4.18 j g • °c . the calorimeter constant was found to be 20.0 j

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10-11-2017
Chemistry

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A student conducts an experiment to determine the enthalpy of solution for lithium chloride dissolved in water. the student combines 5.00 grams of lithium chloride with 100.0 ml of distilled water. the initial temperature of the water is 23.0°c and the highest temperature after mixing reaches 33.0°c. assume a density of 1.00 g/ml and a specific heat of 4.18 j g • °c . the calorimeter constant was found to be 20.0 j

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barnuts barnuts · Answer 1 · 2017-11-21T07:34:21+01:00

I believe that this problem has 6 questions:

a) Is the reaction endothermic or exothermic? Why?
b) Calculate the temperature change of the solution in the calorimeter.
c) Calculate the total mass of the solution in the calorimeter.
d) Calculate the energy change of the solution.
e) Calculate the energy change of the calorimeter.
f) Calculate the total change of enthalpy of the solution in kJ/mole.

The answers are the following:

a) This is exothermic because the temperature of the water increased hence heat energy was released by the reaction.

b) 33 – 23 = 10C

c) 100 ml x 1 g/1 ml = 100 g

d) The formula for change in energy is:

Energy change = mass of solution x temperature change x specific heat
Energy change = 100 g x 10C x 4.18 Joules/gC = 4,180 Joules

e) 20 Joules/C x 10C = 200 Joules

f) First calculate for moles of LiCl: 5.00 g LiCl x 1 mole LiCl/42.4 g = 0.118 moles
So the total energy given off = 4,180 J + 200 Joules = 4,380 Joules
Enthalpy change, delta H is -4,380
-4380 Joules/0.118 moles = -37,119 Joules/mole
-37,119 Joules/ 1mole x 1 kilojoule/1000 Joules = -37.1 kilojoules
ΔH = -37.1 kJ/mole