Question #a4aeb
1 Answer
Here's what I got.
Explanation:
So, you're dealing with an equilibrium reaction between nitrogen tetroxide,
#"N"_2"O"_text(4(g]) rightleftharpoons color(red)(2)"NO"_text(2(g])#
Looking at the equilibrium reaction, it's important to realize what the
For every mole of nitrogen tetroxide that dissociates, the reaction produces
So, you know that you're starting with
For every
This means that out of
#1 color(red)(cancel(color(black)("mole N"_2"O"_4))) * "20 moles dissociate"/(100color(red)(cancel(color(black)("moles N"_2"O"_4)))) = "0.2 moles N"_2"O"_4#
that will dissociate to form nitrogen dioxide. Using the aforementioned mole ratio, this will result in
#0.20color(red)(cancel(color(black)("moles N"_2"O"_4))) * (color(red)(2)" moles NO"_2)/(1color(red)(cancel(color(black)("mole N"_2"O"_4)))) = "0.40 moles NO"_2#
produced by the equilibrium reaction. So, at equilibrium, the reaction vessel will contain
#0.80# moles of#"N"_2"O"_4 -># left undissociated#0.40# moles of#"NO"_2 -># formed by the reaction
Now, the total pressure of the mixture is said to be equal to
In your case, the mole fractions of the two gases will be
#chi_(N_2O_4) = (0.80 color(red)(cancel(color(black)("moles"))))/( (0.80 + 0.40) color(red)(cancel(color(black)("moles")))) = 2/3#
#chi_(NO_2) = (0.40 color(red)(cancel(color(black)("moles"))))/((0.80 + 0.40) color(red)(cancel(color(black)("moles")))) = 1/3#
This means that you have
#P_(N_2O_4) = chi_(N_2O_4) xx P_"total"#
#P_(N_2O_4) = 2/3 * "1 atm" = color(green)(2/3 "atm")#
and
#P(NO_2) = 1/3 * "1 atm" = color(green)(1/3"atm")#
By definition,
#K_p = ( ("NO"_2)^color(red)(2))/(("N"_2"O"_4)) = (1/3)^color(red)(2) "atm"^2* 3/2 "atm"^(-1) = 1/9 * 3/2"atm" = color(green)(1/6" atm")#