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Equilibrium Constant, K, and Reverse Reactions Tutorial

Key Concepts

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Introduction

It is very important to remember that the mass-action expression and therefore the equilibrium law expression for a chemical reaction is based on the chemical reaction as it is written, with the left hand side of the chemical equation defined as reactants and the right hand side of the chemical equation defined as products.

If we are talking about the synthesis of gas C using the gaseous reactants A(g) and B(g) we would write the chemical equation representing this as:

Synthesis of C(g)
reactants products
A(g) + B(g) C(g)

and we talk about the forward reaction being the reaction in which A(g) and B(g) react together to produce the product C(g).

Then we write the expression for the equilibrium constant (K) based on defining the products of the forward reaction as being on the right hand side of the chemical equation:

K =   [C(g)]  
[A(g)][B(g)]

Note that the reaction does not go to completion, rather, while reactants are reacting to produce the product, the product is also decomposing to reform the reactants!

We can write the chemical equation for this reverse reaction, the decomposition of C(g) to produce A(g) and B(g), as:

Decomposition of C(g)
reactants products
C(g) A(g) + B(g)

and now we would talk about the forward reaction being the one in which C(g) is the reactant that decomposes to produce the products A(g) and B(g)

The expression for the equilibrium constant (K′) based on defining the products of the forward reaction as being on the right hand side of the chemical equation:

K′ = [A(g)][B(g)]
  [C(g)]  

Let′s define the forward reaction as the synthesis reaction and the reverse reaction as the decomposition reaction:

  reactants products    
  A(g) + B(g) forward

reverse
C(g)    
forward reaction:
(synthesis of C)
A(g) + B(g) C(g)  
K =   [C(g)]  
[A(g)][B(g)]
reverse reaction:
(decomposition of C)
C(g) A(g) + B(g)  
K′ = [A(g)][B(g)]
  [C(g)]  

Now let′s see what happens when we multiply K by K′:

K × K′ =   [C(g)]  
[A(g)][B(g)]
× [A(g)][B(g)]
  [C(g)]  
= 1

K × K′ = 1 therefore K =   1  
K′
and K′ =   1  
K

In general we can say that for a chemical reaction with equilibrium constant K at a constant temperature T, reversing the chemical reaction at the same temperature T results in an equilibrium constant that is the recripocal of K.

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Worked Example

Question: PCl5(g) decomposes to produce PCl3(g) and Cl2(g) according to the following chemical equation:

PCl5(g) ⇋ PCl3(g) + Cl2(g)

At 250°C the value for the equilibrium constant for this reaction is 0.040

Determine the value of the equilibrium constant for the following reaction at 250°C:

PCl3(g) + Cl2(g) ⇋ PCl5(g)

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

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

Calculate equilibrium constant K (at 250°C) for PCl3(g) + Cl2(g) ⇋ PCl5(g)

PAUSE PAUSE to Prepare a Game Plan
 

  1. What data have you been given?

    PCl5(g) ⇋ PCl3(g) + Cl2(g)     K = 0.040 (at 250°C)

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

    (a) The reaction we have been given is the reverse of the reaction for which we need to calculate the value of K, that is:

    PCl5(g) ⇋ PCl3(g) + Cl2(g) is the reverse of PCl3(g) + Cl2(g) ⇋ PCl5(g)

    (b) At the same temperature:

    Kforward reaction =         1        
    Kreverse reaction

GO GO with the Game Plan
  Determine the value for Kc

PCl5(g) ⇋ PCl3(g) + Cl2(g)     Kforward = 0.040

PCl3(g) + Cl2(g) ⇋ PCl5(g)     K = 1 ÷ Kforward = 1 ÷ 0.040 = 25

PAUSE PAUSE to Ponder Plausibility
  Is your answer plausible?

Consider the value for the forward reaction we were given:

PCl5(g) ⇋ PCl3(g) + Cl2(g)     Kforward = 0.040

K is not large so the equilibrium position must lie to the left, that is most PCl5(g) does not decompose to produce Cl2(g) and PCl3(g) at this temperature, in other words, Cl2(g) and PCl3(g) readily react to produce PCl5(g).

This means that if we write the reverse reaction, the synthesis reaction instead:

PCl3(g) + Cl2(g) ⇋ PCl5(g)

In this reaction, the equilibrium position will lie to the right since we have determined above that Cl2(g) and PCl3(g) readily react to produce PCl5(g).

When we write the expression for the equilibrium constant:

K =   [PCl5(g)]  
[PCl3(g)][Cl2(g)]

the concentration of the product PCl5(g) will be greater than the concentration of the reactants, so we expect K for this synthesis reaction to be greater than K for the decomposition reaction (the original reaction we were given).

Since our calculated value for K is 25, which is larger than K = 0.04 for the original reaction, we are confident our answer is correct.

Note, that if we multiply the two equilibrium constants, their product should equal 1:

Kforward × Kreverse = 1

0.04 × 25 = 1

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

K = 25 for the reaction PCl3(g) + Cl2(g) ⇋ PCl5(g) at 250°C

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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 generalisations regarding the extent of reaction and the magnitude of the equilibrium constant.