Question

Question #60f23

Answer 1

The equilibrium constant for this reaction will be equal to 3.13.

So, you know that you're dealing with a generic equilibrium that involves three species, `A`, `B`, both reactants, and `C`, the product.

Even before doing any calculations, you can predict that `K_c` will be bigger than 1. Notice that the concentration of `A` decreases, while the concentration of `C` increases.

This tells you that the initial concentrations will cause the equilibrium to shift right, i.e. the reaction quotient, `Q_c`, is smaller than `K_c`.

If this is the case, the equilibrium concentration of `B` will decrease as well, but not by the same amount as the concentration of `A` - this happens because of the difference between the stoichiometric coefficients of the two reactants.

So, you were only given two equilibrium concentrations, more specifically those of `A` and of `C`. However, you can use the initial concentrations to determine what the equilibrium concentration of `B` would be.

To do that, use an ICE table for you equilibrium reaction

`" " " "A " "+ " "color(red)(2)B " " rightleftharpoons " "C`
I......0.650............1.20...............0.700
C.......(-x)..............(-`color(red)(2)`x)...................(+x)
E.....0.650-x........1.20-2x..........700+x

You know that `[A]_"equilibrium" = "0.450 M"`. This means that

`0.650 - x = 0.540 => x = 0.200`

Check to see if it matches the equilibrium concentration of `C`

`[C]_"equilibrium" = 0700 + x = 0.700 + 0.200 = "0.900 M"`

This means that the equilibrium concentration of `B` will be

`[B]_"equilibrium" = 1.2 - 2x = 1.2 - 2 * 0.200 = "0.800 M"`

Therefore, the equilibrium constant for this reaction will be

`K_c = ([C])/([A] * [B]^color(red)(2)) = 0.900/(0.450 * 0.800"^2) = color(green)(3.13)`

SIDE NOTE The value of `Q_c` at the start of the reaction was 0.74, which again confirms our initial prediction.