The following system is at equilibrium with [N2O4] = 0.55 M and [NO2] = 0.25 M. If an additional 0.10 M NO2 is added to the reaction mixture with no change in volume, what will be the new equilibrium concentration of N2O4?

     N2O4(g) 2 NO2(g)

UGH! I've been working on this problem for about an hour now for my test tomorrow at WMU. No luck. =( My guess? Impossible. =P

The following system is at equilibrium with [N2O4] = 0.55 M and [NO2] = 0.25 M. If an additional 0.10 M NO2 is added to the reaction mixture with no change in volume, what will be the new equilibrium concentration of N2O4?

N2O4(g) 2 NO2(g)

To find the new equilibrium concentration of N2O4 after adding 0.10 M NO2, we need to apply Le Chatelier's principle. According to Le Chatelier's principle, when a stress is applied to a system at equilibrium, the system adjusts in such a way as to minimize the effect of that stress.

In this case, adding 0.10 M NO2 will increase the concentration of NO2 in the reaction mixture. According to the balanced equation, 1 mole of N2O4 reacts with 2 moles of NO2. So, when additional NO2 is added, it will react with N2O4 to form more NO2.

Let's assume x is the change in concentration of N2O4. Since 1 mole of N2O4 reacts with 2 moles of NO2, the change in NO2 concentration will be 2x, as it reacts in a 1:2 ratio with N2O4.

Now, let's set up an expression to represent the equilibrium concentrations:

[N2O4] = [N2O4(initial)] + x
[NO2] = [NO2(initial)] + 2x

Here, [N2O4(initial)] is the initial concentration of N2O4 before adding NO2, and [NO2(initial)] is the initial concentration of NO2.

Given:
[N2O4(initial)] = 0.55 M
[NO2(initial)] = 0.25 M

We can substitute these initial concentrations into the expressions:

[N2O4] = 0.55 + x
[NO2] = 0.25 + 2x

Now, we need to consider the new concentration of NO2 after adding 0.10 M. The concentration of NO2 will be: [NO2(initial)] + (0.10 M) = 0.25 + 0.10 = 0.35 M.

We can equate the new concentration with the expression we derived earlier:

0.35 = 0.25 + 2x

Now, solve the equation for x:

0.35 - 0.25 = 2x
0.10 = 2x
x = 0.10/2
x = 0.05

Now that we have the value of x, we can find the new equilibrium concentration of N2O4:

[N2O4] = 0.55 + 0.05
[N2O4] = 0.60 M

Therefore, the new equilibrium concentration of N2O4 after adding 0.10 M NO2 is 0.60 M.