I cannot solve this problem. Please help

Question

At equilibrium, the concentrations in this system were found to be  and 

If more  is added, bringing its concentration to  what will the final concentration of  be after equilibrium is re‑established?

Answer 1

Hi Dari,

:-) Nitrogen monoxide is a cardiovascular signaling molecule, plays a role in destruction of the ozone layer, and is a precursor to smog pollution. So, this type of calculation may actually come back to you if you study the environment or the heart.

You are given that the tank is at equilibrium (so inside the tank, the N₂ molecules and O₂ molecules are hitting each other to form NO just as fast as the NO falls apart and converts back to N₂ and O₂.)

Step #1) Find the equilibrium constant

The expression for the equilibrium constant is…

K_c = [NO]² / {[N₂][O₂]}

Since you are given that the tank is at equilibrium, then you can find the value for K_c.

K_c = [0.600]² / {[0.100][0.100]}

K_c = 36

Step #2) Disturb the equilibrium

If you add 0.900 M of the NO, you will then mess up the equilibrium and cause the [NO] to now be 1.000 M, so now, the tank has too much NO. You are not at equilibrium anymore. Since there is too much NO product, the backward reaction will be faster than the forward reaction. In other words, the NO will fall back to N2 and O2 faster than the N2 and O2 react to form product. You will need to set up an I.C.E chart to find the new equilibrium.

Step #3) Find the new equilibrium using an I.C.E. chart

Now there is now too much NO in the tank. Therefore, the backward reaction is dominant. This means that the NO will start to decrease and the reactants will start to increase. So notice the sign of the “Change” values in the I.C.E. chart below. Since the reverse reaction is favored, then you are losing product and gaining reactant.

Reaction: N2 + O2 <=> 2NO

Initial:   0.100M    0.100M    1.000M

Change:   +X        +X          -2X

Equilibrium: (0.1+X) (0.1+X) (1.0-2X)

K_c = [NO]² / {[N₂][O₂]}

36 = [1.0-2X]² / {[0.1+X][0.1+X]}

36 = [1.0-2X]² / [0.1+X]²

take the square root of both sides

6 = (1.0-2X) / (0.1+X)

6(0.1+X) = 1.0-2X

0.6 + 6X = 1 - 2X

4X = 0.4

X = 0.1

Final Answer:

The final concentration of NO is ... [NO] = (1.0 - 2X) = 1.0 - 2(.1) = 0.800 M

Answer 2

Hi Dari R.,

What information and techniques did you attempt to bring to the problem? Can't solve is a rather strong statement; don't know how to solve is closer, and don't want to use my notes is (alas) perhaps closest?

You are given a system at equilibrium, and a chemical reaction that connects the gases. That should clue you in that you should find the equilibrium constant, since that is the single fact that relates all parts of such a system. To write the equilibrium constant, it is = ([product1]^p1 * [product2]^p2 ....)/([reactant1]^r1*[reactant2]^r2 ....) where product is a chemical on the right side of the reaction equation, and reactant is a chemical on the left side of the reaction equation, and the respective exponents (r1, r2, p1, p2 ....) are the numerical coefficients of each chemical in the balanced raction equation, for as many reactants and products as appear in the reaction equation.

So in the present reaction, K = [NO]²/([N₂]*[O₂]) . Put in your initial conditions (molar concentrations before the additional gas was added) and solve for K.

Now, set up an ICE table (initial, change, equilibrium). Express the initial here as [NO]=0.900 , and [O₂] and [N₂] as each 0.100 . Let the change in [NO] be -2x , and the change in [O₂] and [N₂] be each +x . Do you see why it might be easier to calculate using those as changes rather than -x and +x/2, +x/2 respectively? If you don't, then you will shortly!

So your final concentrations are [NO] = 0.9-2x , [O₂] = [N₂] = 0.1 + x each

Put these concentrations into an equilbrium equation. The equilibrium constant K has NOT CHANGED, that's why it's called a constant!

And solve for x (it's a quadratic equation in x) . Plug that value back in to the ICE table values (for E condition) and you have solved for your new equilibrium gases concentration.

Dari, you need to know how to do this type of problem, almost in your sleep. So study up on it, until you can get right down to setting up the ICE table, in less than 1 minute for any new problem. That speed will come with practice.

-- Cheers, -- Mr. d.