Question

Question #c2cfc

Answer 1

`8.08 * 10^(-4)` `"atm"`

Explanation:

You know that when the following reaction takes place at `"500 K"`

`color(blue)(2)"A"_ ((g)) rightleftharpoons "B"_ ((g))`

the equilibrium constant is equal to

`K_p = 2.77 * 10^(-5)`

Even without doing any calculations, the fact that you have `K_p < 1` tells you that you should expect the equilibrium partial pressure of `"B"` to be smaller than the equilibrium partial pressure of `"A"`.

Now, you know that when volume and temperature are kept constant, the pressure of a gas is directly proportional to the number of moles of gas present in the sample.

Notice that it takes `color(blue)(2)` moles of `"A"` to produce `1` mole of `"B"`. This is equivalent to saying that in order for the partial pressure of `"B"` to increase by a value `x`, the partial pressure of `"A"` must decrease by `color(blue)(2)x`.

You can thus say that once equilibrium is reached, you will have

`0 + x = x ->` the equilibrium partial pressure of `"B"`

`5.40 - color(blue)(2)x ->` the equilibrium partial pressure of `"A"`

By definition, the equilibrium constant for this reaction looks like this

`K_p = (P_"B")/(P_"A")^color(blue)(2)`

In your case, you will have

`2.77 * 10^(-5) = x/((5.40 - color(blue)(2)x)^color(blue)(2)`

Now, because the equilibrium constant is significantly smaller than the initial pressure of `"A"`, you can use the approximation

`5.40 - color(blue)(2)x ~~ 5.40`

The above equation becomes

`2.77 * 10^(-5) = x/5.40^color(blue)(2)`

Solve for `x` to find

`x = 2.77 * 10^(-5) * 5.40^color(blue)(2) = 8.08 * 10^(-4)`

Therefore, you can say that the equilibrium partial pressure of `"B"` will be

`P_ "B" = color(darkgreen)(ul(color(black)(8.08 * 10^(-4)color(white)(.)"atm")))`

The answer is rounded to three sig figs, the number of sig figs you have for the initial pressure of gas `"A"`.