# Question c7292

Dec 29, 2017

Here's my take on this.

#### Explanation:

Start by writing the equilibrium constant that describes this equilibrium reaction--keep in mind that you're dealing with the equilibrium partial pressures of water and hydrogen gas here!

${K}_{p} = \left({p}_{\text{H"_ 2))/(p_ ("H"_ 2"O}}\right)$

Now, you know that, at this given temperature, you have

${K}_{p} = 2$

This implies that the equilibrium partial pressure of hydrogen gas is two times higher than the equilibrium partial pressure of water vapor.

${p}_{\text{H"_ 2) = 2 * p_ ("H"_ 2"O")" " " "color(darkorange)("(*)}}$

According to the balanced chemical equation, in order for the reaction to produce $1$ mole of hydrogen gas, it must consume $1$ mole of water vapor.

This is equivalent to saying that in order for the number of moles of hydrogen gas to increase by $1$ mole, the number of moles of water must decrease by $1$ mole.

Since the volume and the temperature at which the reaction takes place are kept constant, you can say that the partial pressures of the two gases will be proportional to the number of moles of each gas.

This means that in order for the partial pressure of hydrogen gas to increase by $\text{1 atm}$, the partial pressure of water vapor must decrease by $\text{1 atm}$.

Let's assume that the initial partial pressure of water vapor is equal to

(p_("H"_2"O"))_0 = p_0color(white)(.)"atm"

If you take ${p}_{{\text{H}}_{2}}$ to be the equilibrium partial pressure hydrogen gas, you can say the equilibrium partial pressure of water vapor is

${p}_{{\text{H"_ 2"O") = p_0 - p_ ("H}}_{2}}$

This means that in order for the partial pressure of hydrogen gas to increase by ${p}_{{\text{H}}_{2}}$, the partial pressure of water vapor must decrease by ${p}_{{\text{H}}_{2}}$.

This is equivalent to

${p}_{\text{H"_ 2) = p_0 - p_ ("H"_ 2"O}}$

Plug this into equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$ to get

${p}_{0} - {p}_{\text{H"_ 2"O") = 2 * p_ ("H"_ 2"O}}$

${p}_{0} = 3 {p}_{\text{H"_ 2"O}}$

This means that you have

${p}_{\text{H"_ 2"O}} = {p}_{0} / 3$

You can thus say that the partial pressure of water vapor decreased by

"% decrease" = (color(red)(cancel(color(black)(p_0))) - color(red)(cancel(color(black)(p_0)))/3)/color(red)(cancel(color(black)(p_0))) xx 100% = 66.7%

This is equivalent to saying that the number of moles of water decreased by 66.7% after the reaction took place. At equilibrium, the number of moles of water vapor is 66.7%# lower than the number of moles of water vapor you started with.