Chemistry Equilibirum constant

Question

Suppose that 0.350 mol of A and 0.520 mol of B are placed a 1.50L container and the following hypothetical equilibrium is established. If the equilibrium amount of C is 0.150 mol, what is the equilibrium constant for this reaction?

What is the equilibrium constant for the reverse reaction?

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

Answer 1

The equilibrium constant K_eq (and a lot of other chemical constants) follow the format of product over reactant. The equation for K_eq for this problem looks like:

K_eq = [C]³/[B]²[A]

Notice that the molar equivalents of the balanced reaction become exponents in the equation.

Now let's look at the balanced equation and think about what's happening. We are told that we end with 0.150 mol of C. Well, our C in our balanced equation has 3 molar equivalents (the 3 in front of the C). We want to know how much A and B we had to use to get there. Think about if I told you 3x = 0.15. How would you solve for x? Like this:

0.150 / 3 = 0.05

now each molar equivalent is 0.05 moles. To solve for A, we need just x, and for B, we need 2x:

1(0.05) = 0.05 mol of component A and

2(0.05) = 0.10 mol of component B

to make the product C. Since we had to use up that amount of A and B to make C, we need to subtract from our initial number of moles:

A: 0.350 - 0.05 = 0.30 mol A

B: 0.520 - 0.10 = 0.42 mol B

And add that amount to our product:

C: 0 + 0.15 = 0.15 mol C

Now we have all of our molar measurements. However, if we notice in our equilibrium constant equation, they dont want moles - they want concentration, i.e. molarity. To solve for molarity, we use:

M = moles/L

We just calculated the number of moles for each, and they've kindly told us the reaction is in 1.5 L. Now we can easily calculate the new concentration of each component:

A = 0.30 / 1.5 = 0.20 M

B = 0.42 / 1.5 = 0.28 M

C = 0.15 / 1.5 = 0.1 M

Now we simply plug our concentrations into our K_eq equation (be mindful of the exponents!)

K_eq = [0.1]³/[0.28]²[0.2] = 6.38*10^-2

But they want the reverse reaction! Meaning

3C ↔ 2B + A

Now our K_eq is flipped. Luckily, we don't have to re-type the bolded work into our calculator again (although you could) - we can simply calculate the inverse of our bolded K_eq:

K_eq(rev) = 1/6.38e-2 = 15.68

It isn't part of the problem, but consider which direction is more favorable! (Hint: it's the reverse reaction)