# Question 1671c

Jul 30, 2017

Here's what I got.

#### Explanation:

You know that at a given temperature, the equilibrium constant that describes this equilibrium

${\text{AB"_ ((g)) rightleftharpoons "A"_ ((g)) + "B}}_{\left(g\right)}$

is equal to $4$ times the total pressure in the reaction vessel once equilibrium is established.

${K}_{p} = 4 \cdot {P}_{\text{total}}$

Now, if you take $x$ to be the number of moles of $\text{AB}$ that react to produce $\text{A}$ and $\text{B}$, and ${\text{AB}}_{0}$ to be the initial number of moles of $\text{AB}$, you can say that, at equilibrium, you will have

• ${\text{AB}}_{0} - x \to$ the number of moles of $\text{AB}$
• $x \to$ the number of moles of $\text{A}$
• $x \to$ the number of moles of $\text{B}$

Assuming that the temperature at which the reaction takes place and the volume of the reaction vessel remain unchanged, you can say that the pressures of the three chemical species are proportional to the number of moles present in the reaction vessel.

This means that you can take ${\text{AB}}_{0} - x$, $x$, and $x$ to be the equilibrium pressures of $\text{AB}$, $\text{A}$, and $\text{B}$, respectively.

Notice that the total number of moles present in the reaction vessel at equilibrium is equal to

overbrace("AB"_ 0 -x)^(color(blue)("moles of AB")) + overbrace(x)^(color(blue)("moles of A")) + overbrace(x)^(color(blue)("moles of B")) = "AB"_0 + x

This means that you can take the total pressure of the mixture to be equal to ${\text{AB}}_{0} + x$.

By definition, the equilibrium constant is equal to

K_p = (P_"A" * P_"B")/P_"AB"

In your case, you will have

${K}_{p} = \frac{x \cdot x}{{\text{AB}}_{0} - x}$

This will be equal to

${x}^{2} / \left({\text{AB"_0 -x) = 4 * ("AB}}_{0} + x\right)$

Rearrange to get

${x}^{2} = 4 \cdot \left({\text{AB"_0 + x) * ("AB}}_{0} - x\right)$

${x}^{2} = 4 \cdot \left({\text{AB}}_{0}^{2} - {x}^{2}\right)$

${x}^{2} = 4 A {\text{B}}_{0}^{2} - 4 {x}^{2}$

$5 {x}^{2} = 4 {\text{AB}}_{0}^{2}$

Finally, you will have

x = sqrt(4/5"AB"_0^2) = (2sqrt(5))/5 * "AB"_0#

Therefore, you can say that, at equilibrium, the reaction will produce

${\text{moles of A" = (2sqrt(5))/5 * "AB}}_{0}$

Here ${\text{AB}}_{0}$ represents the initial number of moles of $\text{AB}$.