Equilibrium Constant

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  • Created by: chunks-42
  • Created on: 07-10-15 08:45

Amounts of Products and Reactants Remain Constant

1. Lots of changes are revserible - they can go both ways. To show a change is reversible, you stick in a <-->.

2. As the reactants get used up, the forward reaction slows down - and as more product is formed, the reverse reaction speeds up. After a while, the forward reaction will be going at exactly the same rate as the backward reaction.

The amounts of reactants and products won't be changing any more, so it'll seem like nothing's happening. It's a bit like your digging a hole while someone else is filling it in at exactly the same speed. This is called a dynamic equilibrium.

3. Equilibrium can be set up in physical state, e.g.

When liquid bromine is shaken in a closed flask, some of it changes to orange bromine gas. After a while, equilibrium is reached - bromine liquid is still changing to bromine gas and bromine gas is still changing to bromine liquid, but they are changing at the same rate.

Br2 (l) <--> Br2 (g)

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Amounts of Products and Reactants Remain Constant

and chemical systems, e.g.

If hydrogen gas and iodine gas are mixed together in a closed flask, hydrogen iodide is formed.

H2 + I2 <--> 2HI

Imagine that 1.0 mole of hydrogen gas is mixed with 1.0 mole of iodine gas at a constant temperature of 640 K. When this mixture reaches equilibrium, there will be 1.6 moles of hydrogen iodide and 0.2 moles of both hydrogen gas and iodine gas. No matter how long you leave them at this temperature, the equilibrium amounts never change. As with the physical system, it's all a matter of the forward and backward rates being equal.

3. A dynamic equilibrium can only happen in a closed system at a constant temperature.

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If you know the molar concentration of each substance at equilibrium, you can work out the equilibrium constant, Kc. Your value of Kc will only be true for that particular temperature.

Before you can calculate Kc, you have to write an expression for it. Here's how:

For the general reaction aA + bB <--> dD + eE

Kc = [D]d x [E]e/[A]a x [B]b

  • The products go on the top line. The square brackets, [], mean concentration in mol dm-3
  • The lower case letters a, b, d and e are the number of moles if each substance.

So for the reaction H2 + I2 <--> 2HI, Kc = [HI]2 / [H2] x [I2]

These little numbers look like the orders of reaction that you saw in rate equations. But they're not - they're the number of moles.

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Calculating Kc

If you know the equilibrium concentrations, just bung them into your expressions. Then with a bit of help from the old calculator, you can work out the value of Kc. The unts are a bit trickier though - they vary, so you have to work them out after each calculation.

E.g. If the volume of the closed flask in the hydrogen iodide example above is 2.0 m3, what is the equilibrium constant temperature at 640 K? The equilibrium concentrations are:

[HI] = 0.8 mol dm3, [H2] = 0.1 mol dm3, and [I2] = 0.1 mol dm3

Just stick the concentrations into the expression for Kc:

Kc = [HI]2 / [H2] x [I2] = [0.8]2 / [0.1] x [0.1] = 64 (Value for Kc)

To wok out the units of Kc put the units in the expression instead of the number:

Kc = (mol dm-3) x (mol dm-3) / (mol dm-3) x (mol dm-3) - the concentration units cancel, so there are no units for c.

So Kc is just 64.

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Calculating Equilibrium Concentrations

You might need to figure out some of the equilibrium concentrations before you can find Kc:

e.g. 0.20 moles of phosphorus (V) chloride decomposes at 600 K in a vessel of 5.00 dm3. The equilibrium mixture is found to contain 0.08 moles of chlorine. Write the expression for Kc and calculate its value, including units.

PCl5 <--> PCl3 + Cl2

First find out how many moles of PCl 5 and PCl3 there are at equilibrium:

The equation tells you that when 1 mole of PCl5 decomposes, 1 mole of PCl3 and 1 mole of Cl2 are formed. So, if 0.08 moles of chlorine are produced at equilibrium, then there will be 0.08 moles of PCl3 as well. 0.08 mol of PCl5 must have decomposed, so there will be 0.12 molles left (0.2 - 0.08).

Divide each number of moles by the volume of the flask to give the molar concentrations:

[PCl3] = [Cl2] = 0.08/ 5.00 = 0.16 moldm-3, [PCl5] = 0.12/5.00 = 0.24 moldm-3

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Calculating Equilibrium Concentrations Continued

Put the concentrations in the expression for Kc and calculate it:

Kc = [PCl3] x [Cl2] / [PCl5] = [0.016] x [0.016] / [0.024] = 0.011

Now find the units for Kc: Kc = (moldm-3)(moldm-3) / moldm-3 = moldm-3

So Kc = 0.011 moldm-3

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Calculating Concentrations in an Equilibrium Mixtu

E.g. When ethanoic acid was allowed to reach equilibrium with ethanol at 25 *C, it was found that the equilibrium mixture contained 2.0 moldm-3 ethanoic acid and 3.5 moldm-3 ethano. The Kc of the equilibrium is 4.0 at 25*C. What are the concentrations of the other components?

CH3COOH + C2H5OH <--> CH3COOC2H5 + H2O

Put all the values you know in the Kc expression :

Kc = [CH3COOC2H5] [H2O] / [CH3COOH] [C2H5OH] --> 4.0 = [CH3COOC2H5] [H2O] / 2.0 X 3.5

Rearranging this gives: [CH3COOC2H5] [H2O] = 4.0 X 2.0 X 3.5 = 28.0

From the equation, you know that [CH3COOC2H5] = [H2O], so: [CH3COOC2H5] = [H2O] = (Square Root)28 = 5.3 moldm-3

The concentration of CH3COOC2H5 and H2O is 5.3 moldm-3

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What happens if the conditions change?

If you change the concentration, pressure or temperature of a reversible reaction, you're going to alter the position of equilibrium. This just means you'll end up with different amounts of reactants and products at equilibrium.

If the position of equilibrium moves to the left, you'll get more reactants

If the position of equilibrium moves the the right, you'll get more products.

There's a rule that lets you predict how the position of equilibrium will change if a condition changes. This rule is known as Le Chatelier's Principle. Here it is:

If there's a change in concentration, pressure or temperature, theequilibrium will move to help counteract the change.

So, basically, if you raise the temperature, the position of equilibrium will try to shift to cool things down. And if you raise the pressure or concentration, the position of equilibrium will shift to try to reduce it.

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What affects Kc?


1. If you increase the temperature, you add heat. The equilibrium shifts in the endothermic (positive delta H) direction to absorb the heat.

2. Decreasing the temperature removes heat. The equilibrium shifts in the exothermic (negative delta H) direction to try to replace the heat.

3. If the forward reaction id endothermic, the reverse reaction will be exothermic and vice versa.

4. If the change means more product is formed, Kc will rise. If it means less product is formed, the Kc will decrease.

The reaction below is exothermic in the forward direction. If you increase the temperature, the equilibrium shifts to the left to absorb the extra heat. This means that less product is formed.

2SO2 + O2 <--> 2SO3 (D.H = -197 kjmol-1)

Kc = [SO3]2 / [SO2]2 x [O2] Since there's less product formed, Kc decreases.

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What affects Kc? Continued


The value of the equilibrium constant, Kc, is fixed at a given temperature. So if the concentration of one thing in the equilibrium mixture changes then the concentrations of the others change to keep the value of Kc the same.

CH3COOH + C2H5OH <--> CH3COOC2H5 + H2O

If you increase the concentration of CH3COOH then the equilibrium will move to the right to get rid of the extra CH3COOH - so more CH3COOC2H5 and H2O are produced. This keeps the equilibrium constant the same.


Catalysts have no effect on the position of equilibrium or no value of Kc. They can't increase the yield - but they do mean equilibrium is appraoched faster.

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