Chemical Equilibrium………..Let’s look at problem #12 in the homework….

The reaction of elemental hydrogen and fluorine to form hydrofluoric acid has an equilibrium constant of 1.15 x 102 at a certain temperature. In one experiment, 3.00 mol of all 3 components, including the product, was added to a 1.50-liter flask. Calculate the equilibrium concentrations of all species.

Write the reaction:

H2 (g) + F2 (g) --------> 2 HF (g)

This is an odd problem, because there is actually some of the

product present before the reaction actually begins! This is not common – product isn’t usually present until the reaction happens!

Chemical Equilibrium……….. We want to set up an ICE:

1 H2 (g) + 1 F2 (g) --------> 2 HF (g)

I

C

E

Remember, I mean initial, C means change, and E means end or equilibrium!

And we use concentration – not moles, or amount!

What are the initial concentrations of each chemical?

Chemical Equilibrium……….. What does our ICE look like?

1 H2 (g) + 1 F2 (g) --------> 2 HF (g)

I 2M 2M 2M

C -x -x +2x

E 2 - x 2 - x 2 + 2x

Kc 115 = [HF]2 = [2 + 2x]2

1 [H2]1[F2]1 [2-x]1[2-x]1

Remember, we insert the concentrations into our K at equilibrium!

We want to solve for the amounts at equilibrium – this is going to require some algebra!

Chemical Equilibrium………..Kc 115 = [HF]2 = [2 + 2x]2

1 [H2]1[F2]1 [2-x][2-x]

We have to expand this…..

115 = 4 + 8x + 4x2

1 4 - 4x + x2

460 – 460x + 115x2 = 4 + 8x + 4x2

Rearranging,

0 = 456 – 468x + 111x2 or 111x2 – 468x + 456 = 0

How do we solve this?

Chemical Equilibrium………..111x2 – 468x + 456 = 0

Use the quadratic formula!

•You can use the program on your calculator to solve this

•You can go to http://www.wolframalpha.com/ and enter in the equation – it will solve it for you!

•What are the solutions for x?

•X = 1.5284 and 2.6878

•One of these solutions can’t work….

•Which one? Why?

Chemical Equilibrium………..

H2 (g) + F2 (g) --------> 2 HF (g)

I 2M 2M 2M

C -x -x +2x

E 2 - x 2 - x 2 + 2x

X = 1.5284 and 2.6878

It is impossible to lose 2.6878 when you only start with 2!

This is the non-real solution –

So x = 1.5284

What are the amounts at equilibrium then?

Chemical Equilibrium………..

H2 (g) + F2 (g) --------> 2 HF (g)

I 2M 2M 2M

C -x -x +2x

E .4716 .4716 5.0568

Let’s plug these values back into K to see if they equal 115/1!

Kc 115 = [HF]2 = [5.0568]2 114.975/1

1 [H2]1[F2]1 [.4716]1[.4716]1

We can set up a ratio of pressures instead of concentrations We can set up a ratio of pressures instead of concentrations too!too!

This is called KThis is called Kpp

Let’s look at number 24 – Let’s look at number 24 – Hydrogen gas and iodine vapor react to form hydrogen Hydrogen gas and iodine vapor react to form hydrogen

iodide gas. The equilibrium constant, or Kiodide gas. The equilibrium constant, or Kcc, is 1.00 x 10, is 1.00 x 1022

at 25at 2500C. Suppose hydrogen at 5.0 x 10C. Suppose hydrogen at 5.0 x 10-1-1 atm, iodine at 1.0 atm, iodine at 1.0 x 10x 10-1 -1 atm, and hydrogen iodide at 5.0 x 10atm, and hydrogen iodide at 5.0 x 10-1-1atm are added atm are added to a 5.0 L flask. Calculate the equilibrium pressures of all to a 5.0 L flask. Calculate the equilibrium pressures of all species.species.

What is the reaction?What is the reaction?

Hydrogen gas and iodine vapor react to form hydrogen Hydrogen gas and iodine vapor react to form hydrogen iodide gas. The equilibrium constant, or Kiodide gas. The equilibrium constant, or Kcc, is 1.00 x 10, is 1.00 x 1022

at 25at 2500C. Suppose hydrogen at 5.0 x 10C. Suppose hydrogen at 5.0 x 10-1-1 atm, iodine at 1.0 atm, iodine at 1.0 x 10x 10-1 -1 atm, and hydrogen iodide at 5.0 x 10atm, and hydrogen iodide at 5.0 x 10-1-1atm are added atm are added to a 5.0 L flask. Calculate the equilibrium pressures of all to a 5.0 L flask. Calculate the equilibrium pressures of all species.species.

HH2 (g)2 (g) + I + I2 (g) 2 (g) --------> 2 HI --------> 2 HI (g)(g)

What would KWhat would Kpp look like? look like?

KKpp = [P HI] = [P HI]22

[P H[P H22]]11[P I[P I22]]11

Where P is the pressure, instead of concentration, of the gas!Where P is the pressure, instead of concentration, of the gas!

But the problem gives us the KBut the problem gives us the Kcc, and not the K, and not the Kpp – are they the – are they the

same? No!same? No!

KKpp = K = Kcc (RT) (RT)nn

What is delta n?What is delta n? It is the moles of gas produced minus the moles of gas you It is the moles of gas produced minus the moles of gas you

start with – start with – In this reaction, you make 2 moles of HI, and start with In this reaction, you make 2 moles of HI, and start with

one mole of Hone mole of H22, and one mole of I, and one mole of I22

So, So, n = 2 – 2 = 0n = 2 – 2 = 0 KKpp = 100 [(.0821)(298)] = 100 [(.0821)(298)]00

KKpp = 100 = 100

In this case, the KIn this case, the Kpp actually did equal the K actually did equal the Kcc!!

So…..So…..

H2 (g) + I2 (g) --------> 2 HI (g)

I .5 atm .1 atm .5 atm

C -x -x +2x

E .5 – x .1 – x .5 + 2x

Same exact setup as before – but we are using pressures instead of concentrations, so we use a Kp and not a Kc!

Kp = [.5+2x]2 = 100

[.5-x]1[.1-x]1

Can you do the algebra?

Try and do problems 1-31 for homework Try and do problems 1-31 for homework this weekend….this weekend….

I will do a homework check on Tuesday….I will do a homework check on Tuesday…. Remember, if concentrations are given, use Remember, if concentrations are given, use

a Ka Kcc – if pressures are given, use a K – if pressures are given, use a Kpp – –

To convert between them:To convert between them: KKpp = K = Kcc(RT)(RT)nn