Question #b2748

The first thing that you need to do here is to figure out the equilibrium constant of the reaction by using the known equilibrium concentrations of the three chemical species.

#"N"_ (2(g)) + "O"_ (2(g)) rightleftharpoons color(red)(2)"NO"_ ((g))#

By definition, the equilibrium constant for this equilibrium looks like this

#K_c = (["NO"]^color(red)(2))/(["N"_2] * ["O"_2])#

You know that at a given temperature, the system is at equilibrium when

#["N"_2] = ["O"_2] = "0.100 M "# and #" " ["NO"] = "0.500 M"#

Even without doing any calculations, you should be able to say that #K_c > 1# because the reaction vessel contains more product than reactants at equilibrium, which implies that at this temperature, the forward reaction is favored.

So, plug in the values to find

#K_c = (0.500^color(red)(2))/(0.100 * 0.100) = 25#

Now, you are told that the concentration of nitric oxide is being increased to #"0.800 M"#.

In response, the system will react by changing the position of the equilibrium in such a way as to reduce the added stress, i.e. to decrease the concentration of nitric oxide #-># think Le Chatelier's Principle here.

This means that the reverse reaction will be favored. The concentration of nitric oxide will decrease by a value #x# #"M"#, which implies that the concentrations of nitrogen gas and oxygen gas will increase by #(x/color(red)(2))# #"M"# #->#this is due to the #1:color(red)(2)# mole ratio that exists between the two reactants and nitric oxide.

You can thus say that equilibrium constant will be equal to--keep in mind that the equilibrium constant is, well, constant as long as the temperature of the reaction remains constant!

#K_c = (0.800 - x)^color(red)(2)/((0.100 + x/color(red)(2)) * (0.100 + x/color(red)(2)))#

#K_c = (0.800 -x )^color(red)(2)/(((0.200 + x)/color(red)(2)) * ((0.200 + x)/color(red)(2))#

#K_c= (4 * (0.800 - x)^color(red)(2))/(0.200 + x)^2#

This will be equivalent to

#sqrt(K_c) = sqrt((4 * (0.800 - x)^color(red)(2))/(0.200 + x)^2)#

#sqrt(K_c) = (2 * (0.800 - x))/(0.200 + x)#

which will give you

#(0.200 + x) * sqrt(25) = 2 * (0.800 - x)#

#1 + 5x = 1.600 - 2x#

#7x = 0.600 implies x = 0.600/7 = 0.0857#

Therefore, you can say that when the equilibrium is reestablished, the concentration of nitric oxide will be

#["NO"] = "0.800 M" - "0.0857 M" = color(darkgreen)(ul(color(black)("0.714 M")))#

Notice that the final operation is a subtraction, so you must round the answer to three decimal places, the number of decimal places you have for #"0.800 M"#.