Chemical Equilibrium

Question

At equilibrium, a 1.0 liter container was found to contain 0.20 moles of A, 0.20 moles of B, 0.40 moles of C and 0.40 mole of D. If 0.10 moles of A and 0.10 moles of B are added to this system,

A(g) + B(g) C(g) + D(g)

what will be the new equilibrium concentration of A?

Answer 1

A(g) + B(g) <==> C(g) + D(g)

0.2......0.2............0.4.......0.4.......Equilibrium

FInd K:

K = [C][D] / [A][B] = (0.4)(0.4) / (0.2)(0.2) = 4

After addition of 0.1 mols A and B, equilibrium will shift to the left (more products)...

A(g) + B(g) <==> C(g) + D(g)

0.3......0.3..............0.4.........0.4......Initial

0.3-x...0.3-x...........0.4+x.....0.4+x....Change

0.3-x...0.3-x...........0.4+x.....0.4+x...Equilibrium

K = 4 = [C][D] / [A][B] = (0.4-x)² / (0.3+x)²

Solve for x and subtract that from 0.3 to find new equilibrium concentration of A.

Answer 2

Hello,

You are starting out at equilibrium, so let's calculate the equilibrium constant:

Concentrations are as follows (just take the moles of each species and divide by the volume):

A - 0.2M, B - 0.2M, C - 0.4M and D - 0.4M.

K = Products/reactants = [C]*[D]/[A]*[B] = 4.

After the addition of A and B the new concentrations are 0.3M for A and 0.3M for B.

Make an Ice Chart:

A + B -> C + D

Initial 0.3 | 0.3 0.4 | 0.4

Change -x | -x +x | +x

End 0.3-x | 0.3-x 0.4+x | 0.4+x

Set up an equation to find x based on the end state (at the end you are at equilibrium)

K = [C]*[D]/[A]*[B]

4 = (0.4+x)(0.4+x)/(0.3-x)(0.3-x)

solve for x and you get

x=1 and x=0.066667.

x can't be 1 because then the concentrations of A and B will be negative.

so x=0.066667 and concentration of A is 0.3-x so it is 0.23M.

Checking the answer here:

https://www.chem.tamu.edu/class/fyp/mcquest/ch17.html

Question 14, the answer is showing as 0.23M at the answer key.