When Compounds do not React in a 1:1 Molar Ratio

• The molar ratios affect how the concentrations change during a reaction. • We must account for these ratios when we solve ICE box problems. • Specifically they will affect the CHANGE line when we are doing these problems.

Practice Problem Nitrosyl bromide (NOBr) decomposes

according to the reaction below:

2 NOBr 2 NO + Br2

0.500 M of NOBr is used initially and when the reaction reaches equilibrium, 0.410 M NOBr remain. Determine the equilibrium constant! (Keq)

Start with the Initial Line!• We know the starting concentration of NOBr.• What about the concentrations of our products?• Do we know any other concentrations?

NOBr NO Br2

Initial 0.500 0 0

Change

Equilibrium 0.410

Next is the Change Line!2 NOBr 2 NO + 1 Br2

• We don’t know the exact change amount yet, but we know how they are going to change. Pay attention to the coefficients!

NOBr NO Br2

Initial 0.500 0 0

Change - 2 X 2 X X

Equilibrium 0.410

• Using the info for NOBr, we can solve for X, the change in concentration:

0.500 M – 2x = 0.410 M X = 0.045 M

• But remember this is just X. • We are looking for the

equilibrium constant (K) so we need the rest of the values at equilibrium..

Solving for XNOBr

Initial 0.500

Change - 2 X

Equilibrium 0.410

Solving the Concentrations at Equilibrium• X = 0.045 M (Previous Slide)

• Note: These numbers are plugged into the equation

NOBr NO Br2

Initial 0.500 M 0 0

Change

Equilibrium 0.410 M 0.090 M 0.045 M

- 2 X 2 X X

2 NOBr 2 NO + Br2

• Remember coefficients when writing the equilibrium expression.

K = [NO]2[Br2] [NOBr]2

K = [0.090]2[0.045] [0.410]2

K = 0.0021 or 2.1x10-3

NOBr NO Br2

Equilibrium 0.410 0.090 0.045

• Plug in the values at equilibrium into the expression and solve for K:

Practice Problem – A different kind…• Hydrogen gas reacts with Iodine gas to produce

hydroiodic acid. Write and balance the chemical equation:

1 H2 + 1 I2 2 HI• Calculate all concentrations at equilibrium when

initially 0.200 M of both reactants are used and the equilibrium constant is equal to 64.0.

• What makes this problem different than previous ones? – We are given the Keq value and asked to solve for the concentrations.

Start with the Initial!• Again, this one is different because we are not

given any concentrations at equilibrium. We are going to solve for that upcoming.

H2 l2 Hl

Initial 0.200 0.200 0

Change

Equilibrium ?? ?? ??

Next Comes the Change!

1 H2 + 1 I2 2 HI• Make sure you’ve balanced the equation first!

H2 l2 Hl

Initial 0.200 0.200 0

Change -X -X 2 X

Equilibrium

Finally the Equilibrium!• To fill in the rest of the boxes, it is always:

INITIAL + CHANGE = EQUILIBRIUM• Great! Now it’s time for us to solve for X !

H2 l2 Hl

Initial 0.200 0.200 0

Change -X -X 2 X

Equilibrium 0.200 – X 0.200 – X 2 X

Wait a second…• There is no column for us to solve for X!

• Fear not my downhearted disciples of chemistry! We were given the Keq intially in the problem.

• We will write equilibrium expression and use the Keq to solve for X. Then plug that in to get the concentrations at equilibrium.

H2 l2 HlEquilibrium 0.200 – X 0.200 – X 2 X

1 H2 + 1 I2 2 HI• Remember coefficients when

writing the equilibrium expression.

K = [HI]2

[H2][I2]

64.0 = [2X]2

[0.200 – X][0.200 – X]

H2 l2 Hl

Equilibrium 0.200 – X 0.200 – X 2 X

• Plug in the values at equilibrium, and Keq= 64.0 into the expression and solve for X:

Let’s clean this up a little bit…• In the bottom, two identical things are

multiplied together. That means we can ???

• So what can we do to get X by itself? • Take the square root!

8.0 = 2X 0.200 – X

64.0 = [2X]2

[0.200 – X]2

Let’s clean this up a little bit…

• Need X, so multiply both sides by???

8.0 = 2 X 0.200 – X

8.0 (0.200 – X) = 2 X *Distribute Through*1.6 – 8 X = 2 X *Combine Like Terms*1.6 = 10 X *Solve for X!*

0.16 = XBut remember that’s not the final answer.That’s just X, now lets plug it into the

expression.

Yes, That is My Final Answer!

H2 l2 Hl0.200 – X 0.200 – X 2 X

• From the last slide, we solved for X = 0.16. Lets Plug it in!

• And that’s it! At equilibrium there will be[0.040] H2, [0.040] I2, [0.32] HI

0.200 – 0.160.040 0.200 – 0.160.040 2(0.16)0.32