At 430^@"C"430∘C, an equilibrium mixture consists of 0.0200.020 moles of "O"_2O2, 0.0400.040 moles of "NO"NO, and 0.960.96 moles of "NO"_2NO2. Calculate KpKp for the reaction, given the total pressure is "0.20 atm"0.20 atm ?
2"NO"(g) + "O"_2(g) rightleftharpoons 2"NO"_2(g)2NO(g)+O2(g)⇌2NO2(g) .
I know that the answer is 1.5 * 10^51.5⋅105 atm with significant figures, but I am not sure how to get this. Help is much appreciated!
I know that the answer is
1 Answer
Explanation:
You know that
2"NO"_ ((g)) + "O"_ (2(g)) rightleftharpoons 2"NO"_ (2(g))2NO(g)+O2(g)⇌2NO2(g)
At
"0.020 moles O"_20.020 moles O2 "0.040 moles NO"0.040 moles NO "0.96 moles NO"_20.96 moles NO2
Even without doing any calculations, you should be able to predict that
K_p " >> " 1Kp >> 1
because the equilibrium mixture contains significantly more product than reactants, which implies that the equilibrium lies to the right, i.e. the forward reaction is favored at this temperature.
Now, at equilibrium, the total pressure inside the reaction vessel is equal to
As you know, the partial pressures of the three gases will depend on their respective mole ratios, as described by Dalton's Law of Partial Pressures.
For nitrogen oxide, you will have
P_ "NO" = chi_ "NO" * P_"total"PNO=χNO⋅Ptotal
Here
chi_ "NO" = (0.040 color(red)(cancel(color(black)("moles"))))/((0.040 + 0.020 + 0.96)color(red)(cancel(color(black)("moles")))) = 0.040/1.02
represents the mole fraction of nitrogen oxide in the reaction vessel at equilibrium.
Therefore,
P_ "NO" = 0.040/1.02 * "0.20 atm"
Use the same approach to calculate the partial pressure of oxygen gas and nitrogen dioxide. For oxygen gas, you will have
P_ ("O"_ 2) = (0.020 color(red)(cancel(color(black)("moles"))))/((0.040 + 0.020 + 0.96)color(red)(cancel(color(black)("moles")))) * "0.20 atm"
P_ ("O"_ 2)= 0.020/1.02 * "0.20 atm"
Similarly, for nitrogen dioxide, you will have
P_ ("NO"_ 2) = (0.96 color(red)(cancel(color(black)("moles"))))/((0.040 + 0.020 + 0.96)color(red)(cancel(color(black)("moles")))) * "0.20 atm"
P_ ("NO"_ 2) = 0.96/1.02 * "0.20 atm"
The equilibrium constant for this reaction can be written using the partial pressures of the three gases
K_p = (P_ ("NO"_ 2)^2)/( P_ "NO"^2 * P_ ("O"_ 2))
Plug in your values to find
K_p = ( (0.96/1.02 * 0.20)^2 color(red)(cancel(color(black)("atm"^2))))/( (0.040/1.20 * 0.20)^2 color(red)(cancel(color(black)("atm"^2))) * 0.020/1.02 * "0.20 atm")
K_p = (0.96^2 * color(red)(cancel(color(black)((0.20/1.02)^2))) * 1.02)/(0.040^2 * color(red)(cancel(color(black)((0.20/1.02)^2))) * 0.020 * "0.20 atm")
K_p = (0.96^2 * 1.02)/(0.040^2 * 0.020 * "0.20 atm") = color(darkgreen)(ul(color(black)(1.5 * 10^5color(white)(.)"atm"^(-1))))
The answer is rounded to two sig figs.
