A mixture containing 0.0392 M A(g) and 0.0452 M B9g) is allowed to come to equilibrium at 300K.

The reaction: 3A(g)+ 2B(g)=> C(g)+ D(g) occurs. At equilibrium, [C]=0.00128 M
What is the value of Kc?

To find the value of Kc, we need to write the balanced equation for the reaction and express the equilibrium concentrations of the reactants and products in terms of x, where x is the change in concentration.

The balanced equation for the reaction is: 3A(g) + 2B(g) -> C(g) + D(g)

Let's assume that at equilibrium, the concentrations of A, B, C, and D are [A], [B], [C], and [D], respectively.

According to the reaction stoichiometry, at equilibrium, the concentration of A would be [A] - 3x, the concentration of B would be [B] - 2x, and the concentration of C would be [C] + x.

Since the given information is [C] = 0.00128 M, we can write the equilibrium expression (Kc):

Kc = [C]^c [D]^d / [A]^a [B]^b

Since the coefficients of A, B, C, and D in the balanced equation are 3, 2, 1, and 1, respectively, the equilibrium expression becomes:

Kc = ([C] + x) * ([D] + x) / ([A] - 3x)^3 * ([B] - 2x)^2

Substituting the given values into the equation:

Kc = (0.00128 + x)(x) / (0.0392 - 3x)^3 * (0.0452 - 2x)^2

Now, we can find the value of x by substituting the given value of [C] into the equation:

0.00128 = (0.00128 + x)(x) / (0.0392 - 3x)^3 * (0.0452 - 2x)^2

Simplifying the equation and solving for x yields:

x = 0.00128 * (0.0392 - 3x)^3 * (0.0452 - 2x)^2 / (0.00128 + x)

x ≈ 0.00115 M

Finally, substitute the value of x into the equilibrium expression to find the value of Kc:

Kc = (0.00128 + 0.00115)(0.00115) / (0.0392 - 3 * 0.00115)^3 * (0.0452 - 2 * 0.00115)^2

Kc ≈ 3.54 * 10^3 (rounded to three significant figures)

To find the value of Kc, we first need to understand the relationship between the concentrations of the reactants and products at equilibrium. The equilibrium expression for the given reaction is:

Kc = [C]^p[D]^q / ([A]^x[B]^y)

where [C], [D], [A], and [B] represent the molar concentrations of C, D, A, and B respectively, and p, q, x, and y are the stoichiometric coefficients of C, D, A, and B in the balanced equation.

In this case, the balanced equation is:

3A(g) + 2B(g) -> C(g) + D(g)

From the balanced equation, we can see that p = 1 (since there is one C), q = 1 (since there is one D), x = 3 (since there are three As), and y = 2 (since there are two Bs).

We are given that at equilibrium [C] = 0.00128 M. Therefore, we can substitute this value into the equilibrium expression:

Kc = (0.00128)^1(0)^1 / (0.0392)^3(0.0452)^2

Note that the concentration of D is not given, so we assume it to be 0 for now.

Kc = 0.00128 / (0.0392)^3(0.0452)^2

Now we need to calculate the value of Kc using the given concentrations of A and B. Substituting the values into the expression:

Kc = 0.00128 / (0.0392)^3(0.0452)^2

Kc = 0.00128 / (0.0392)^3(0.0452)^2

Kc = 0.00128 / (6.62 x 10^-5)(9.17 x 10^-5)

Kc = 23.13 (approx)

Therefore, the value of Kc is approximately 23.13.

3A + 2B ==> C + D

Set up an ICE chart.
Initial:
A = 0.0392 M
B = 0.0452 M
C = 0 M
D = 0 M

part of equilibrium:
C = 0.00128 M

change:
Therefore, C must have changed by + 0.00128.
D must have changed by + 0.00128
A must have changed by -3*0.00128
B must have changed by -2*0.00128

remainder of equilibrium:
A is now 0.0392-change in A.
B is now 0.0452-change in B.
C is 0.00128
D is 0.00128

Now set up Keq expression, plug in the value and solve for Kc.