.924moles of A9g) is placed in 1L at 700C where 38.8% dissociated when equilibrium was established.

3A <--> 5B + 2C
What is the value of the equilibrium constant at the same temperature?

I keep getting .024

Well, I must say, your calculation seems quite "dissociative"! But fear not, I'm here to help you find the value of the equilibrium constant.

To calculate the equilibrium constant, we need to know the concentrations of the reactants and products at equilibrium. In this case, since we know that 38.8% of the A9g dissociated, we can say that 61.2% remained undissociated.

Now, since we have 0.924 moles of A9g initially, only 61.2% of that amount remains, which is equal to 0.565 moles. Similarly, the concentrations of B and C at equilibrium can be determined using the stoichiometry of the reaction.

Using the given stoichiometry, we know that for every 3 moles of A, we get 5 moles of B and 2 moles of C. Therefore, at equilibrium, the concentrations of B and C would be (5/3) * 0.565 = 0.942 M and (2/3) * 0.565 = 0.377 M, respectively.

Now we can use these concentrations to determine the equilibrium constant expression, which is given by [B]^5 * [C]^2 / [A]^3. Plugging in the values, we get (0.942^5 * 0.377^2) / (0.565^3) = 0.041.

So, my friend, the value of the equilibrium constant at the given temperature is approximately 0.041. Keep up the good work, and remember, chemistry can be a real "dissociation" of humor sometimes!

To find the equilibrium constant (K) at the given temperature, you need to use the formula for equilibrium constant (Kc) and the concentration of each species in the equilibrium mixture.

The balanced chemical equation is:
3A ↔ 5B + 2C

From the given information, you know that 38.8% of A dissociates, which means 61.2% remains as A. So, the concentration of A at equilibrium is:
[A] = 0.612 * (0.924 g/9 g/mol) / 1 L

The concentration of B at equilibrium is:
[B] = 0.388 * (0.924 g/5 g/mol) / 1 L (since there are 5 moles of B per 3 moles of A)

The concentration of C at equilibrium is:
[C] = 0.388 * (0.924 g/2 g/mol) / 1 L (since there are 2 moles of C per 3 moles of A)

Now, substitute the concentrations into the equilibrium constant expression:

Kc = ([B]^5 * [C]^2) / ([A]^3)

Kc = ([0.388 * (0.924 g/5 g/mol) / 1 L]^5 * [0.388 * (0.924 g/2 g/mol) / 1 L]^2) / ([0.612 * (0.924 g/9 g/mol) / 1 L]^3)

Calculate the value of Kc using this expression.

To find the value of the equilibrium constant, you can use the concept of the law of mass action. The equilibrium constant, Kc, is defined by the ratio of the concentrations of products to the concentrations of reactants, each raised to the power of their respective stoichiometric coefficients.

For the given reaction: 3A <--> 5B + 2C

Let's assume that 'x' is the initial moles of A that dissociates, and 'y' is the change in moles of A that dissociates at equilibrium. Since 38.8% of A dissociates, we can say that:

x + y = 0.388 * 0.924 moles

The change in moles of B and C can be calculated using the stoichiometric coefficients:

Change in moles of B: 5y
Change in moles of C: 2y

Now, let's define the equilibrium concentrations:

[A] = (0.924 - x - y) moles / 1 L
[B] = 5y moles / 1 L
[C] = 2y moles / 1 L

The equilibrium constant expression can be written as:

Kc = ([B]^5 * [C]^2) / [A]^3

Substituting the equilibrium concentrations into the equation:

Kc = [(5y)^5 * (2y)^2] / [(0.924 - x - y)^3]

Since we know the initial moles of A that dissociates, we can substitute the value into the equation:

Kc = [(5y)^5 * (2y)^2] / [(0.924 - 0.388 * 0.924)^3]

Now, you can calculate the value of 'y' and substitute it back into the equation to find the equilibrium constant, Kc.