Exercise 6.102 ) 90,10 : > Constarts 1 BeriodicTable PartA Dry ice in solid carb

Question

Exercise 6.102 ) 90,10 : > Constarts 1 BeriodicTable PartA Dry ice in solid carbon deside, Instead ofl melting sold carbon eside sublimen according to the eovation CO()-CO,(s).When dry ie is added to wam water, hea Express your answer usling three significent igures hyrfas 0ooler holds 15.0 L of water heated oM'C Part Celculate the mans of dry lce that should be added to the water so that he dry lce coplely sublimes away when the water reaches 11 C. Assume no heat Express yeur answer sing twe significant igues Provide Feecback

Explanation / Answer

Answers: A. 33.90 kJ/mole, B. 5950.68 g

Explanation: A) Change in Enthalpy = [the sum of all enthalpy of formation of all products] - [the sum of all enthalpy of formation of all reactants]

CO2(s) == CO2(g)
From a table the Hf of CO2(g) = -393.5 kJ/mole

Enthalpy of dry ice sublimation = 1(-393.5 kJ/mole ) – (1(-427.4 kJ/mole)
-393.5 + 427.4 = 33.90 kJ/mole

B)q = mcT

where q is the heat change, m the mass of the substance water, c the specific heat capacity of water and T the change in temperature.

c = 4.186 J/g/K
m = 15000 g (as 1 mL of water = 1 g at ambient conditions)
T = (11 - 84) K = -73 K
15,000 g x 73 deg x 4.186 J/g/K = 4583670 J (endothermic reaction)
Now you can calculate q.

The next step is to calculate how much dry ice is required to cause that amount of heat change. This is really simple (dimensional analysis should tell you right away how to do this) given H(sublimation) = 33.9 kJ/mol:

n(dry ice) = q / H(sublimation)

4583.670 kJ / 33.9 kJ/mol = 135.212 mol
135.212 mol x 44.01 g/mol = 5950.68 g

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