# Question 9ef8a

Mar 17, 2017

Here;s what I got.

#### Explanation:

You can't really calculate the value of ${K}_{p}$ because you're missing the temperature at which the reaction takes place.

You can calculate the value of ${K}_{c}$, but you must know the temperature at which the reaction takes place in order to be able to calculate the value of ${K}_{p}$.

So, for the equilibrium reaction

$2 {\text{SO"_ (2(g)) + "O"_ (2(g)) rightleftharpoons 2"SO}}_{3 \left(g\right)}$

you have

${K}_{c} = \left(\left[{\text{SO"_3]^2)/( ["SO"_2]^2 * ["O}}_{2}\right]\right)$

As you know, the expression of ${K}_{c}$ uses the equilibrium concentrations of the three chemical species.

In your case, you will have

${K}_{c} = \left(\text{0.259 M")^2/( ("0.59 M")^2 * "0.05 M}\right)$

K_c = (0.259^2 color(red)(cancel(color(black)(("mol L"^(-1))^2))))/(0.59^2 color(red)(cancel(color(black)(("mol L"^(-1))^2))) * "0.05 mol L"^(-1))

${K}_{c} = \text{3.854 mol"^(-1)"L}$

Now, you should know that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{K}_{p} = {K}_{c} \cdot {\left(R T\right)}^{\Delta n}}}}$

Here

• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
• $T$ is the absolute temperature at which the reaction takes place
• $\Delta n$ is the difference between the total number of moles on the products' side and the total number of moles on the reactants' side

• $\text{2 moles SO"_2 + "1 mole O"_2 = "3 moles gas}$

You have two moles of sulfur dioxide reacting with one mole of oxygen gas on the reactants' side

• $\text{2 moles of SO"_3 = "2 moles of gas}$

You have two moles of sulfur trioxide being produced on the products' side

This means that

$\Delta n = {n}_{\text{total products" - n_"total reactants}}$

will be equal to

$\Delta n = 2 - 3 = - 1$

You can now say that ${K}_{p}$ will be equal to

${K}_{p} = {K}_{c} \cdot {\left(R T\right)}^{- 1}$

If you take $T$ $\text{K}$ to be the temperature at which the reaction takes place, you can say that

K_p = "3.854 mol"^(-1)"L" * (0.0821 ("atm" * "L")/("mol" * color(red)(cancel(color(black)("K")))) * Tcolor(white)(.) color(red)(cancel(color(black)("K"))))^(-1)

K_p = 3.854 color(red)(cancel(color(black)("mol"^(-1)))) color(red)(cancel(color(black)("L"))) * 1/(0.0821 * T) color(red)(cancel(color(black)("mol")))/("atm" * color(red)(cancel(color(black)("L"))))

which gets you

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{K}_{p} = \left(\frac{46.9}{T}\right) \textcolor{w h i t e}{.} {\text{atm}}^{- 1}}}}$

All you have to do to get the actual value of ${K}_{p}$ is to plug in the value you have for the absolute temperature at which the reaction takes place.

Now, does this result make sense?

Notice that in this case, ${K}_{p}$ is equal to

${K}_{p} = \left(\left({\text{SO"_3)^2)/(("SO"_2)^2 * ("O}}_{2}\right)\right)$

Keep in mind that the expression for ${K}_{p}$ uses the equilibrium partial pressures of the three gases.

If you measure the partial pressures of the three gases in atmospheres, you will have -- using only units

K_p = ( color(red)(cancel(color(black)("atm"^2))))/(color(red)(cancel(color(black)("atm"^2))) * "atm") = "atm"^(-1)#

This means that, at least from a dimensional point of view, the answer makes sense.