[NewtoAtoms]

Q:  One mole of NOCl gas is added to a 4.0L container. The NOCl undergoes slight decomposition to form NO gas and Cl₂ gas. If the equilibrium constant Kc = 2.0 x 10^-10, calculate the concentrations of all the species at equilibrium at this temperature.

2NOCl (g) <----------> 2NO (g) + Cl₂ (g)

I have done all my leg work and equations, but I desperately need some help...
Could someone be so kind to walk me through this.


STEP #1
                          NOCl                   2NO                    Cl2
Initial                   0.25 M                 O                        O
change                  -x                      +x                      +x
-------------------------------------------------------------------
Equilibrium            0.25 - x               2X                         X

STEP #2

Kc = [NO]²[Cl₂] / [NoCl]²

2.0 x 10^-10 = (x)² (x) / [0.25 - x]²

2.0 x 10^-10 = (x)² (x) / 0.0625 - 0.5 x + x²

2.0 x 10^-10(x² - 0.5x + 0.0625) = (4X²)(x)

2.0 x 10^-10x² - 1.0 x 10-10 x + 1.25 x 10-11 = 4x³

BUT I can't apply the quadradic equation to this because there is X³, x², x, #...

Can anyone help me with this?
Where have I gone so desperately wrong?

[Yggdrasil]

The equilibrium constant for the reaction is very small, which means that only a small amount of product will be formed.  Hence, x will be very small.  This means that 0.25 - x will be approximately equal to 0.25.  If you use this approximation, you can solve the problem.

[NewtoAtoms]

Hello Yggdrasil.

I have a few follow-up questions:

1.  If Kc is a VERY small number then can I ALWAYS follow this reasoning? (ie. 0.25 - x = 0.25)

2.  If 0.25 -x = 0.25 then it's safe to assume that x = 0. Then if I calculate the concentration of all the species at equilibrium I would propose that NOCL = 0.25 M, whereas NO = 0 and Cl₂= O BECAUSE in my Initial-Change-Equilibrium chart I stated that NO = +x and Cl₂ = +x.  I just find that I am not answering the question at all, with hard copy numbers. 

Is this okay in the world of chemists?

I am so grateful for your time and explanations.

[NewtoAtoms]

Or should I just equate 0.25-x = 0.25 back into the equilibrium equation:

Kc = (x)² (x) / (0.25-x)²

2.0 x 10^-10(0.0625) = x³

1.25 x 10^-11 = x³ - CUBE ROOT OTHER SIDE

x = 2.32 x 10^-4

therefore to answer the original question:

1.  the concentration of NOCl at equilibrium would be (0.25 - 2.32 x 10^-4) = 0.249
2.  the concentration of NO at equilibrium would be (2x) = 4.64 x 10^-4
3.  the concentration of Cl2 at equilibrium would be x = 2.32 x 10^-4

[Yggdrasil]

Correct.  I should note that it is possible to solve a cubic equation (formulas exist but they are very messy).  Using a computer program (Mathematica) to solve the cubic equation that you wrote in the OP, I get that x = 0.000232.  This is essentially the same as the answer found using the approximation.

Note that there is only a 0.09% error assuming that [NOCl] is equal to 0.25 whereas there is an 100% error assuming that [NO]=[Cl₂] = 0.  x can be ignored only if it is being added to or subtracted from a sufficiently large number.

These types of approximations are very common in chemistry and physics and are often useful in the derivation of many important formulas.

[Borek]

1.  If Kc is a VERY small number then can I ALWAYS follow this reasoning? (ie. 0.25 - x = 0.25)

You can always follow this reasoning when x << 0.25 (much less). Kc value is irrelevant.

[Yggdrasil]

Except for the fact that a small Kc value indicates that x will be a very small number.  If Kc were 1, then we would not be justified in assuming that x << 0.25.

[Borek]

Perhaps I wasn't clear. What I was aiming at was that whenever x << y we can try to simplify calculations assuming y - x = y. Doesn't matter what the x and y values are.