Question

Question #2c380

Answer 1

The equilibrium constant for this reaction will be `K_"c" = 3.13`.

So, you've got a generic equilibrium reaction and you're asked to determine the equilibrium constant, `K_c`. In order to do that, you need to know the equilibrium concentration of all the species involved in the reaction.

Since you only know the equilibrium concentrations of `"A"` and `"C"`, you can use the initial concentrations to determine what you're missing - the equilibrium concentration of `"C"`.

Before doing any actual calculations, you can predict that `K_c` will be greater than 1, since the equilibrium concentration of the product increases and the equilibrium concentration of the `A` decreases - compared with their initial cocnentrations.

Use an ICE table (more here: https://en.wikipedia.org/wiki/RICE_chart) to find a relationship between the initial and the equilibrium concentrations of all the species

`" "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.............0.700+x

By definition, the equilibrium constant will be

`K_c = (0.700 + x)/((0.650 - x) * (1.20 - 2x)^2)`

You know that

`0.650 - x = 0.450 => x = 0.200`

(you get the same result if you use 0.700 + x = 0.900)

Replace the value of `x` into the equation and you'll get

`K_c = (0.700 + 0.200)/((0.650 - 0.200) * (1.20 - 2*0.200)^2)`

`K_c = (0.900)/(0.450 * 0.800^2) = 3.125`

Rounded to three sig figs, the answer will be

`K_c = color(green)(3.13)`