2.00 mil of A and B are placed in a 500ml container at equilibrium C was found to be 0.16mol/L has formed. Calculate the equilibrium constant for formation of C.

10 months ago

Answered By Tara W

1A + 3B <—> 2C

Step 1.) First calculate the initial concentrations of A and B:

concentration [A]i= molesA /total volume(L)

[A]i=2.00 mol/0.500L = 4.00 mol/L

[B]i=2.00 mol/0.500L = 4.00 mol/L

Step 2.) Next put these values into your ICE table, along with 0mol/L for initial concentration of C:

A B C

I: 4.00M 4.00M 0 M

C:

E:

Step 3.) Deteremine the amount each changes, given by + or - x multiplied by the molar coefficient.

If the reaction is going forwards, the reactants will be -x and the products will be +x, if going backwards, the reactants will be +x and products -x. This is a forward reaction as they are asking for the formation of C, therefore:

change of A = - molar coefficient of A *x= -1x

change of B = -3*x = -3x

change of C = +2*x = +2x

A B C

I: 4.00M 4.00M 0 M

C: -x -3x +2x

E:

Then fill in the equilibrium expression for each in your ICE table with:

[initial concentration]-/+change

A B C

I: 4.00M 4.00M 0 M

C: -x -3x +2x

E: 4.00M-x 4.00M-3x 0M+2x

Step 4.) Next use the equilibrium concentration for C, 0.16 M, and the equilibrium expression for C to solve for x:

[C]_{eq}=0+2x

0.16 M = 2x

0.08 M = x

Now sub this value for x into the equilibrium expressions for A and B to determine their concentrations at equilibrium:

[A]_{eq}=4.00 M - 0.08 M

[A]_{eq}= 3.92 M

[B]_{eq}= 4.00 M - 3*0.08 M

[B]_{eq }= 3.76 M

Step 6.) Now, finally, put the equilibrium concentrations for each into the equilibrium constant expression:

10 months ago

## Answered By Tara W

1A + 3B <—> 2C

Step 1.) First calculate the initial concentrations of A and B:

concentration [A]i= molesA /total volume(L)

[A]i=2.00 mol/0.500L = 4.00 mol/L

[B]i=2.00 mol/0.500L = 4.00 mol/L

Step 2.) Next put these values into your ICE table, along with 0mol/L for initial concentration of C:

A B C

I: 4.00M 4.00M 0 M

C:

E:

Step 3.) Deteremine the amount each changes, given by + or - x multiplied by the molar coefficient.

If the reaction is going forwards, the reactants will be -x and the products will be +x, if going backwards, the reactants will be +x and products -x. This is a forward reaction as they are asking for the formation of C, therefore:

change of A = - molar coefficient of A *x= -1x

change of B = -3*x = -3x

change of C = +2*x = +2x

A B C

I: 4.00M 4.00M 0 M

C: -x -3x +2x

E:

Then fill in the equilibrium expression for each in your ICE table with:

[initial concentration]-/+change

A B C

I: 4.00M 4.00M 0 M

C: -x -3x +2x

E: 4.00M-x 4.00M-3x 0M+2x

Step 4.) Next use the equilibrium concentration for C, 0.16 M, and the equilibrium expression for C to solve for x:

[C]

_{eq}=0+2x0.16 M = 2x

0.08 M = x

Now sub this value for x into the equilibrium expressions for A and B to determine their concentrations at equilibrium:

[A]

_{eq}=4.00 M - 0.08 M[A]

_{eq}= 3.92 M[B]

_{eq}= 4.00 M - 3*0.08 M[B]

_{eq }= 3.76 MStep 6.) Now, finally, put the equilibrium concentrations for each into the equilibrium constant expression:

For, 1A+3B <—> 2C :

k

_{eq}= [C]_{eq}^{2 }/ [A]_{eq}*[B]_{eq}^{3}k

_{eq}= [0.16M]^{2}/ ( [3.92M]*[3.76M]^{3 })k

_{eq}= 1.23 x10^{-4 }