Question #a2c21

The equilibrium concentration of #CO# will be #8.6 * 10^(-3)"M"#.

Before actually doing any calculations, take a second to look at the equilibrium constant for this reaction, #K_c#. Notice that #K_c# is much smaller than one, which means that the equilibrium will lie far to the left, so to speak.

In other words, at equilibrium, the concentrations of the products, #CO# and #O_2#, will be very small in comparison to the concentration of the reactant, #CO_2#.

To actually solve this problem you'll have to use the ICE Chart method (more here: http://en.wikipedia.org/wiki/RICE_chart).

Calculate the initial concentration of the reactant by using the number of moles and the volume of the vessel given

#C = n/V = "2 moles"/"5 L" = "0.4 mol/L"#

Now use the ICE table - remember that each species' concentration changes proportional to their respective stoichiometric coefficients

....#color(red)(2)CO_(2(g)) rightleftharpoons color(blue)(2)CO_((g)) + O_(2(g))#
I.....0.4.....................0..............0
C...#color(red)("-2")x#...................#color(blue)("+2")x#.........+x
E...(0.4-2x).............2x...............x

Use the definition of the equilibrium constant to solve for #x#.

#K_c = ([CO]^(2) * [O_2])/([CO_2]^(2)) = ((2x)^(2) * x)/(0.4-2x)^(2) = (4x^(3))/(0.4 - 2x)^(2) #

Because #K_c# is so small, you can neglect the #2x# term in the denominator and simply the calculations significantly. This will get you

#K_c = (4x^(3))/0.4^(2) = 2.0 * 10^(-6)#

The value of #x# will be #x = 0.004308#, which means that the equilibrium concentration of #CO# will be

#[CO] = 2 * x = 2 * 0.004308 = "0.008616 M"#

Rounded to two sig figs, although it would be better if the answer was rounded to one sig fig - this is because of the number of sig figs in 2 moles and 5 liters, the answer will be

#[CO] = color(red)(8.6 * 10^(-3)"M")#

SIDE NOTE If you don't neglect the 2x term in the denominator, the value of x will be 0.00425, which will make the equilibrium concentration of CO equal to #8.5 * 10^(-3)"M"#.