Question

0.2 moles of A and 0.4 moles of the B were reacted at certain temperature and were allowed to come to the equilibrium the mixture contain 0.1 mole of A find kc if the volume of the container is equilibrium 2.0 dm3

Answer 1

Hey... Here is your answer ^_^

Answer:

Kc=13.4

Explanation:

I'll show my explanation in steps for your understanding

1. First set balance the equation

2A+2B=2AB

2. Now write the equilibrium constant for the reaction

Kc=[AB][A][B]

3. As Kc is being determined the temperature matters and thus volume also matters.So the conversion of moles to molarity is necessary

M=molesL

4. As the volume of the container is 2dm3

Conversionof dm3 to litres is necessary

1dm3=1L

∴ 2dm3=2L

5. Calculate the molarities

Conc. of A = 0.2moles2L=0.1M

Conc. of B = 0.4mole2L=0.2M

Conc. of A at equilibrium 0.1moles2L=0.05M

5. Set up an ICE table

For problems related to equilibrium
concentrations,you need to set up an ICE table.

mmmmml2Alm+mmmm2Bllll→ml2AB
I/mol:mll0.1Mmmmmml0.2Mmmmmml0
C/mol:m−xmmmmmm−xmmmm+2x
E/mol:lm0.1M−xmml0.2M−xlmmml2x

From the given information we know that the equilibrium amount of A is 0.05M. Thus 0.1M−x=0.05M

Solve for x

0.1M−x=0.05M
x=0.1M−0.05M
x=0.05M

6. Solve for the equilibrium amounts of B and AB

AB=2x=2×0.05M=0.1M
B=0.2M−0.05M=0.15M
And we know the equilibrium amounts of A

7. Now solve for Kc

Kc=productsreactants

Kc=[0.1M][0.05M][0.15M]

Kc=13.333

If you want three sig figs

Kc=13.4

You can also calculate the Qc to see to which direction will the reaction proceed

Qc=[AB]2[A]2[B]2

Qc=177.777777778

As Kc<Qc

Reaction will proceed towards left...

Hope it helped you out ⭐^_^⭐
Thanks⭐(^^)⭐