 # Equilibrium Constant, K, and Stoichiometry Tutorial

## Key Concepts

• K is the symbol given to the equilibrium constant for a chemical reaction.1
• The value of K for a given chemical reaction at a particular temperature depends on how the chemical equation is balanced.
For example, the reaction in which A(g) reacts with B(g) to produce C(g) can be balanced in lots of ways using different stoichiometric coefficients (mole ratios):

(1) A(g) + B(g) ⇋ C(g)    K(1) = [C(g)]   [A(g)][B(g)]
(2) 2A(g) + 2B(g) ⇋ 2C(g)    K(2) = [C(g)]2   [A(g)]2[B(g)]2

• There is a relationship between these different values for the equilibrium constant for the same reaction under the same conditions.
For the example given above:

K1 = √K2

K2 = K12

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## Introduction: How Balancing a Chemical Equation Effects the Value of the Equilibrium Constant, K

Imagine we add some hydrogen iodide gas, HI(g), to a 1 L sealed vessel heated to a constant temperature.
We find that the hydrogen iodide gas decomposes, producing hydrogen gas, H2(g), and iodine gas, I2(g).
After a little while the system reaches equilibrium, so that the concentrations of hydrogen iodide gas, hydrogen gas and iodine gas are no longer changing.
The concentration of each species at equilibrium is measured.
The results of the experiment are shown in the table below:

 reactant ⇋ products 3.531 × 10-3 4.789 × 10-4 4.789 × 10-4

The system is at equilibrium, so, by writing the mass-action expression2and evaluating it using the concentrations of each species at equilibrium given above, we could obtain a value for the equilibrium constant, Kc, for this reaction at this temperature.

In order to write the mass-action expression, we first need a balanced chemical equation to represent the equilibrium system.
There are many possible ways we might do this, for example:

• HI(g) ⇋ ½H2(g) + ½I2(g)
• 2HI(g) ⇋ H2(g) + I2(g)

Which balanced equation we use determines the mass-action expression we will write:

chemical equation mass-action expression
(1) HI(g) ⇋ ½H2(g) + ½I2(g)
 Q(1) = [H2(g)]½[I2(g)]½ [HI(g)]
(2) 2HI(g) ⇋ H2(g) + I2(g)
 Q(2) = [H2(g)][I2(g)] [HI(g)]2

Does it make any difference which one we use?
Let's use the equilibrium concentrations given above to calculate the value for Q (which is now the equilibrium constant Kc since the system is at equilibrium):

(1) HI(g) ⇋ ½H2(g) + ½I2(g)
mass-action expression equilibrium constant
 Q(1) = [H2(g)]½[I2(g)]½ [HI(g)]
 K(1) = [4.789 × 10-4]½[4.789 × 10-4]½ [3.531 × 10-3] = 0.136
(2) 2HI(g) ⇋ H2(g) + I2(g)
mass-action expression equilibrium constant
 Q(2) = [H2(g)][I2(g)] [HI(g)]2
 K(2) = [4.789 × 10-4][4.789 × 10-4] [3.531 × 10-3]2 = 0.0184

So, yes, it does matter very much how we choose to balance the chemical equation, but, the values of the two equilibrium constants, K(1) and K(2), are related to each other.
Can you see that Q(2) is the square of Q(1), hence K(2) is the square of K(1).

Q(2) = Q(1) × Q(1) = Q(1)2
 [H2(g)][I2(g)][HI(g)]2
=
 [H2(g)]½[I2(g)]½ [HI(g)] × [H2(g)]½[I2(g)]½ [HI(g)]
=
 ( [H2(g)]½[I2(g)]½ [HI(g)] )2
K(2) = K(1) × K(1) = K(1)2
0.0184 = 0.136 × 0.136 = 0.1362 = 0.0184

And can you see that Q(1) is the square root of Q(2), hence K(1) is the square root of K(2)?

Q(1) = √Q(2)
 [H2(g)]½[I2(g)]½ [HI(g)]
=
 √( [H2(g)][I2(g)] [HI(g)]2 )
K(1) = √K(2)
0.136 = √0.0184 = 0.136

The value of Kc for a particular chemical reaction at a given temperature depends on how you balance the chemical equation, that is, it depends on the stoichiometric coefficient (mole ratios) for each reactant and product species.
It also means that given Kc and the relevant balanced chemical equation, you can derive a different value for Kc based on a different way to balance the chemical equation.

Do you know this?

Play the game now!

## Worked Example

Question: Ammonia gas can be produced when nitrogen gas reacts with hydrogen gas according to the following balanced chemical equation:

N2(g) + 3H2(g) ⇋ 2NH3(g)

At 400°C the value of the equilibrium constant for this reaction is 39.
Calculate the value for the equilibrium constant at 400°C for the following reaction:

½N2(g) + 1½H2(g) ⇋ NH3(g)

Solution: (based on the StoPGoPS approach to problem solving)

STOP STOP! State the Question.
What is the question asking you to do?

Calculate the value of the equilibrium constant, K.

PAUSE PAUSE to Prepare a Game Plan

1. What data have you been given?

(1) N2(g) + 3H2(g) ⇋ 2NH3(g)     K(1) = 39 (at 400°C)

(2) ½N2(g) + 1½H2(g) ⇋ NH3(g)     K(2) = ? (at 400°C)

2. What is the relationship between what you have been given and what you need to find?

(1) N2(g) + 3H2(g) ⇋ 2NH3(g)
 K(1) = [NH3(g)]2     [N2(g)][H2(g)]3
(2) ½N2(g) + 1½H2(g) ⇋ NH3(g)
 K(2) = [NH3(g)]     [N2(g)]½[H2(g)]3/2

K1 = K22
K2 = √K1

GO GO with the Game Plan
Subsitute the value given for K(1) into the equation:

K2 = √K1
K2 = √39 = 6.24

PAUSE PAUSE to Ponder Plausibility

Work backwards: is

K(1) = K(2)2 using our calculated value for K2 ?
K(1) = 6.242 = 39
which is the same as the value we were given so we are reasonably confident that our calculation is correct.

STOP STOP! State the Solution
State your solution to the problem.

At 400°C for the reaction :

½N2(g) + 1½H2(g) ⇋ NH3(g)

Kc = 6.24

Do you understand this?

Take the test now!

1. Although this discussion will concern Kc, the same logic can be applied to KP, for this reason we have used K rather than Kc for the following generalisations.

2. The mass-action expression, Q, is also known as the reaction quotient, or as the concentration fraction.