# Question c2cfc

Jun 29, 2017

$8.08 \cdot {10}^{- 4}$ $\text{atm}$

#### Explanation:

You know that when the following reaction takes place at $\text{500 K}$

$\textcolor{b l u e}{2} {\text{A"_ ((g)) rightleftharpoons "B}}_{\left(g\right)}$

the equilibrium constant is equal to

${K}_{p} = 2.77 \cdot {10}^{- 5}$

Even without doing any calculations, the fact that you have ${K}_{p} < 1$ tells you that you should expect the equilibrium partial pressure of $\text{B}$ to be smaller than the equilibrium partial pressure of $\text{A}$.

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 $\textcolor{b l u e}{2}$ moles of $\text{A}$ to produce $1$ mole of $\text{B}$. This is equivalent to saying that in order for the partial pressure of $\text{B}$ to increase by a value $x$, the partial pressure of $\text{A}$ must decrease by $\textcolor{b l u e}{2} x$.

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

$0 + x = x \to$ the equilibrium partial pressure of $\text{B}$

$5.40 - \textcolor{b l u e}{2} x \to$ the equilibrium partial pressure of $\text{A}$

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

${K}_{p} = {\left({P}_{\text{B")/(P_"A}}\right)}^{\textcolor{b l u e}{2}}$

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 $\text{A}$, you can use the approximation

$5.40 - \textcolor{b l u e}{2} x \approx 5.40$

The above equation becomes

$2.77 \cdot {10}^{- 5} = \frac{x}{5.40} ^ \textcolor{b l u e}{2}$

Solve for $x$ to find

$x = 2.77 \cdot {10}^{- 5} \cdot {5.40}^{\textcolor{b l u e}{2}} = 8.08 \cdot {10}^{- 4}$

Therefore, you can say that the equilibrium partial pressure of $\text{B}$ 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 $\text{A}$.