chemistry question: (equilibrium) please help! thanks so much... 1. for the reac

Question

chemistry question: (equilibrium) please help! thanks so much...

1. for the reaction 2CO(g) <-->C(s)+CO2(g), equilibrium constant (Kc=7.7).

At a particular time, all the substances had 0.034 mol in a 1 L container.

a) what is the Q at these concentrations?

2. At 623 K, the equilibrium constant (Kp) for the reaction: CaCO3(s) <--> CaO(s) + CO2(g) was found to be 0.15. The equilibrium concentration of CO2 at this temperature is 2.93x10^-3 M. If 200g of CaCO3(g) was placed in a 10L vessel and heated to the same temperature and equilibrium was achieved, how many grams of CaCO3(s) are left?

Explanation / Answer

1) we know that

Q = [CO2] / [CO]^2

{ we consider concentration of only gaseous components in the expression, the concentrations of liquids and solids do not change, so they are excluded from the expression )

Now [CO2] = 0.034 / 1 = 0.034 M

[CO]^2 = (0.034 / 1)2 = 0.034 X 0.034

Q = 0.034 / 0.034 X 0.034 = 29.411.

2) We know that we consider concentration of only gaseous components in the expression, the concentrations of liquids and solids do not change, so they are excluded from the expression

so for following reaction

Kp = pCO2

Here Kp = Kc (RT)n   ; n =1 ( number of moles of gaseous products - number of moles of gaseous reactants)

R = 0.0821

T = 623 K

Kc = Kp / RT = 0.15 / 0.0821 X 623 = 0.00293

So equilibrium concentration of CO2 = 0.00293

so moles of CO2 in 10L = 0.00293 X volume = 0.0293 moles

As per the reaction 1 moles of CO2 will be produced from 1 moles of CaCO3

so 0.0293 will be produced from 0.0293 moles

Iniitial moles of CaCO3 = mass / molecular weight = 200 / 100 = 2 moles

so moles of CaCO3 left = 2 -0.0293 = 1.9707 moles

Grams of 1.9707 moles = 1.9707 X molecular weight = 1.9707 X 100 = 197.07 grams will be left

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