Question #b2748

1 Answer
Jul 8, 2017

Answer:

#["NO"] = "0.714 M"#

Explanation:

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"#.