Question #1671c
1 Answer
Here's what I got.
Explanation:
You know that at a given temperature, the equilibrium constant that describes this equilibrium
"AB"_ ((g)) rightleftharpoons "A"_ ((g)) + "B"_ ((g))AB(g)⇌A(g)+B(g)
is equal to
K_p = 4 * P_"total"Kp=4⋅Ptotal
Now, if you take
"AB"_0 - x ->AB0−x→ the number of moles of"AB"AB x ->x→ the number of moles of"A"A x ->x→ the number of moles of"B"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
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
By definition, the equilibrium constant is equal to
K_p = (P_"A" * P_"B")/P_"AB"
In your case, you will have
K_p = (x * x)/("AB"_0 - x)
This will be equal to
x^2/("AB"_0 -x) = 4 * ("AB"_0 + x)
Rearrange to get
x^2 = 4 * ("AB"_0 + x) * ("AB"_0 - x)
x^2 = 4 * ("AB"_0^2 - x^2)
x^2 = 4A"B"_0 ^ 2 - 4x^2
5x^2 = 4"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
"moles of A" = (2sqrt(5))/5 * "AB"_0
Here
