Question

Calcium carbonate (chalk: CaCO3) decomposes according to the following equation:

CaCO3(s) --> CaO(s) + CO2(g)

Part 1) Which of the following represents the correct equilibrium constant expression for this reaction?

A) K = [CaCO3]/([CaO][CO2])

B) K = [CaO][CO2]/[CaCO3]

C) K = [CO2]

Part 2) In a 1.00 L container, there is 0.440 grams of CO2, 1.79 moles of CaCO3 and 0.178 moles of CaO. What is the value of Q at this point?

Part 3) If K for this reaction at a given temperature is 0.0075, which of the following represents how this reaction will proceed? Use the reaction in part 1 and your work in part 2.

A) The reaction will shift toward products, reducing the concentration of reactants

B) The reaction will shift toward reactants, reducing the concentration of products

C) The reaction has reached equilibrium and concentrations of products and reactants will remain stable

Answer 1

1)
Kc is defined as concentration of product by concentration of reactant
Solid do npt appear in Kc expression
Answer: Kc = [CO2]
Answer: C

2)
Lets find the molarity of CO2

Molar mass of CO2,
MM = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol


mass(CO2)= 0.440 g

number of mol of CO2,
n = mass of CO2/molar mass of CO2
=(0.44 g)/(44.01 g/mol)
= 9.998*10^-3 mol
volume , V = 1.00 L


Molarity,
M = number of mol / volume in L
= 9.998*10^-3/1
= 9.998*10^-3 M

Qc = [CO2]
= 9.998*10^-3
= 0.009998
AnsweR: 0.009998

3)
K = 0.0075 < Q
Since K is less than Q, the reaction will shift to reactant side to decrease product
Answer: B