What are the partial pressures of #"NO"_2# and #"Cl"_2# at equilibrium?

At 305 K, this reaction takes place:

#2"NO"_2 (g) + "Cl"_2 (g) rightleftharpoons 2"NO"_2"Cl" (g)#

4.00 atmospheres of chlorine gas and 2.00 atm of nitro dioxide are injected into the reactor, a reaction has taken place and a chemical equilibrium has occurred. The equilibrium of the nitril chloride (product) partial pressure is 0.707 atm.

What are the partial pressures of #"NO"_2# and #"Cl"_2# at equilibrium?

1 Answer
Dec 29, 2017

Answer:

Here's what I got.

Explanation:

The idea here is that the equilibrium partial pressures of the three gases will be proportional to the number of moles of each gas present in the reaction vessel at equilibrium--this is the case because the temperature and the volume at which the reaction takes place are constant.

#2"NO"_ (2(g)) + "Cl"_ (2(g)) rightleftharpoons 2"NO"_ 2"Cl"_ ((g))#

The balanced chemical equation that describes this equilibrium tells you that in order for the reaction to produce #2# moles of nitryl chloride, it must consume #2# moles of nitrogen dioxide and #1# mole of chlorine gas.

This is equivalent to saying that in order for the number of moles of nitryl chloride to increase by #2# moles, the number of moles of nitrogen dioxide must decrease by #2# moles and the number of moles of chlorine gas must decrease by #1# mole.

Since the partial pressure of each gas is proportional to the number of moles, you can say that in order for the partial pressure of nitryl chloride to increase by #"2 atm"#, the partial pressure of nitrogen dioxide must decrease by #"2 atm"# and the partial pressure of chlorine gas must decrease by #"1 atm"#.

You know that at the start of the reaction, the partial pressure of nitryl chloride is equal to #"0 atm"#.

This means that in order for the equilibrium partial pressure of nitryl chloride to be equal to #"0.707 atm"#, the initial partial pressure of nitrogen dioxide must decrease by

#0.707 color(red)(cancel(color(black)("atm NO"_2"Cl"))) * "2 atm NO"_2/(2color(red)(cancel(color(black)("atm NO"_2"Cl")))) = "0.707 atm"#

and the initial partial pressure of the chlorine gas must decrease by

#0.707 color(red)(cancel(color(black)("atm NO"_2"Cl"))) * "1 atm Cl"_2/(2color(red)(cancel(color(black)("atm NO"_2"Cl")))) = "0.3535 atm"#

This means that, at equilibrium, the partial pressures of the three gases will be

#p_ ("NO"_ 2"Cl") = "0.707 atm"#

#p_ ("NO"_ 2) = "2.00 atm" - "0.707 atm" = "1.29 atm"#

#p_ ("Cl"_ 2) = "4.00 atm" - "0.3535 atm" = "3.65 atm"#

The partial pressures of nitrogen dioxide and of chlorine gas are rounded to two decimal places because that's how many decimal places you have for their initial partial pressures.

If you want, you can calculate the equilibrium constant in terms of the partial pressures of the three gases

#K_p = ((p_ ("NO"_ 2"Cl"))^2)/((p_ ("NO"_ 2))^2 * p_ ("Cl"_ 2))#

You will end up with

#K_p = (0.707^2 color(red)(cancel(color(black)("atm"^2))))/(1.29^2 color(red)(cancel(color(black)("atm"^2))) * "3.65 atm") = "0.0823 atm"^(-1)#