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:

For, 1A+3B <—> 2C : 

k_eq = [C]_eq² / [A]_eq*[B]_eq³

k_eq = [0.16M]² / ( [3.92M]*[3.76M]³ )

k_eq = 1.23 x10^-4