# Question #c2cfc

##### 1 Answer

#### Answer:

#### Explanation:

You know that when the following reaction takes place at

#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 **smaller** than the equilibrium partial pressure of

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 **moles** of **mole** of **increase** by a value **decrease** by

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

#0 + x = x -># theequilibrium partial pressureof#"B"#

#5.40 - color(blue)(2)x -># theequilibrium partial pressureof#"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

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

Therefore, you can say that the equilibrium partial pressure of

#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