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Q vs K : Predicting Direction of Reaction Chemistry Tutorial

Key Concepts

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Introduction

At 350°C nitrogen gas (N2(g)) and hydrogen gas (H2(g)) will react to produce ammonia gas (NH3(g)).
This is an equilibrium reaction which we can represent by the following balanced chemical equation:

N2(g) + 3H2(g) ⇋ 2NH3(g)

Imagine an experiment in which we add 1.00 moles of N2(g) and 1.00 moles of H2(g) to a 1.00 L sealed vessel heated to 350°C and then monitor the production of NH3(g) and the consumption of N2(g) and H2(g) over time by measuring the concentration of each of species in the vessel.
The results of the experiment are shown in the table and graph below:

Time [N2(g)] [H2(g)] [NH3(g)]
0 1.00 1.00 0
1 0.874 0.634 0.252
2 0.814 0.422 0.372
3 0.786 0.358 0.428
4 0.781 0.343 0.438
5 0.781 0.343 0.438
concentration
mol L-1		 N2(g) + 3H2(g) ⇋ 2NH3(g)

Time

As we would expect, we see the concentration of the reactants, N2(g) and H2(g), decrease while the concentration of the product, NH3(g), increases, until equilibrium at around time=4 is reached at which point the concentration of each species does not change.
What is happening to the value of the mass-action expression, Q, (or reaction quotient) as the reaction proceeds in the forward direction towards equilibrium?
The value of the mass-action expression at each time has been calculated and is shown in the table and graph below:

Time [N2(g)] [H2(g)] [NH3(g)]
Q =     [NH3(g)]2    
[N2(g)][H2(g)]3
0 1.00 1.00 0 0
1 0.874 0.634 0.252 0.285
2 0.814 0.422 0.372 1.97
3 0.786 0.358 0.428 5.08
4 0.781 0.343 0.438 6.09
5 0.781 0.343 0.438 6.09
Q N2(g) + 3H2(g) ⇋ 2NH3(g)

Time

As the reaction proceeds in the forward direction towards equilibrium, the value of the mass-action expression, Q, is increasing.
At equilibrium, around t=4, the value of the mass-action expression (reaction quotient), Q, becomes constant and is equal to the value of the equilibrium constant, K, for this reaction at this temperature.

At equilibrium: Q = 6.09 = K

Another way to say this is that as long as the value of Q is less than the value of K then the reaction proceeds in the forward direction:

If Q < K
Forward reaction is favoured.
Reaction proceeds in the forward direction.

When Q = K the reaction is at equilibrium, neither forward nor reverse reaction is favoured.

Now imagine that at t=6 we suddenly add 1.00 moles of NH3(g) to this equilibrium mixture above at 350°C and then monitor the concentration of each species over time.
The results of this experiment are shown in the table and graph below:

Time [N2(g)] [H2(g)] [NH3(g)]
5 0.781 0.343 0.438
6 0.781 0.343 1.438
7 0.841 0.523 1.318
8 0.871 0.613 1.258
9 0.881 0.643 1.238
10 0.885 0.655 1.231
11 0.885 0.655 1.231
concentration
mol L-1		 N2(g) + 3H2(g) ⇋ 2NH3(g)

Time

As predicted by Le Chatelier's Principle, adding more NH3(g) to the system that was at equilibrium will drive the reaction in the reverse direction, consuming some of the additional NH3(g) to produce more N2(g) and more H2(g) until equiliubrium is re-established somewhere around t=10.

We can now calculate and graph the value for the mass-action expression at each time:

Time [N2(g)] [H2(g)] [NH3(g)]
Q =     [NH3(g)]2    
[N2(g)][H2(g)]3
5 0.781 0.343 0.438 6.09 = K
6 0.781 0.343 1.438 65.6
7 0.841 0.523 1.318 14.4
8 0.871 0.613 1.258 7.89
9 0.881 0.643 1.238 6.54
10 0.885 0.655 1.231 6.09 = K
11 0.885 0.655 1.231 6.09 = K
Q N2(g) + 3H2(g) ⇋ 2NH3(g)

Time

Notice that the value of Q increases as soon as some more of the "product" NH3(g) is added to the equilibrium mixture.
As the reverse reaction proceeds, consuming NH3(g) and producing N2(g) and H2(g), the value of Q is high but decreasing, until eventually Q = K when equilibrium is re-established.
For as long as the value of Q is greater than K, the reaction will proceed in the reverse direction until equilibrium is re-established at which time Q = K again.

If Q > K
Reverse reaction is favoured.
Reaction proceeds in the reverse direction.

When Q = K the reaction is at equilibrium, neither forward nor reverse reaction is favoured.

Given the chemical equation, the value of the equilibrium constant (K), and the concentrations of each species at a given time, we can determine the direction the reaction is proceeding in just by calculating the value of Q and comparing it to K:

Q < K forward reaction is favoured
Q = K neither forward nor reverse reaction is favoured
(system is at equilibrium)
Q > K reverse reaction is favoured

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Worked Example: Q > K

Question:

For the reaction: N2(g) + O2(g) ⇋ 2NO(g)
the equilibrium constant, Kc, is 1.0 × 10-5 at 1500 K.
Predict the direction the reaction will move in if the reactants and products have the following concentrations:
[N2] = 0.05 mol L-1
[O2] = 0.02 mol L-1
[NO] = 0.30 mol L-1

Solution:

(Based on the StoPGoPS method for problem solving.)

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

Predict the direction the reaction will proceed in order to attain equilibrium.

PAUSE PAUSE to Prepare a Game Plan
 
  1. What data have you been given?

    Reaction: N2(g) + O2(g) &$8651; 2NO(g)
    Initial Concentrations:
    (mol L-1)    		 0.05   0.02   0.30
    Kc = 1.0 × 10-5 at 1500 K

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

    (i) Q =   [NOg]2  
    [N2(g)][O2(g)]

    (ii)
    Q < K Reaction proceeds in the forward direction
    Q = K System is at equilibrium.
    Rate of forward reaction = rate of reverse reaction
    Q > K Reaction proceeds in the reverse direction

GO GO with the Game Plan
  (i) Substitute the values for the concentration of each species into the mass-action expression, Q:
Q =   [NO]2  
[N2][O2]
Q =   [0.30]2  
[0.05][0.02]
Q = 90

(ii) Compare the value of the mass-action expression, Q, with the given value for the equilibrium constant, Kc.

Q = 90
Kc = 1.0 × 10-5

Q > K therefore reaction proceeds in the reverse direction.

PAUSE PAUSE to Ponder Plausibility
  Is your answer plausible?

  1. Check your orders of magnitude
    Simplifiy the values for concentration by rounding to the nearest 10 in order to quickly decide if you have the correct order of magnitude:
    mass-action expression:Q =  [NO]2  
    [N2][O2]    		  
    Calculation we did above:Q =   [0.30]2  
    [0.05][0.02]    		 = 90
    Calculation using approximations:Q ≈   [10-1]2  
    [10-2][10-2]    		 =   10-2  
    10-4    		 = 102 = 100

    Since our approximation for Q (100) is about the same as our calculated value for Q (90) we are confident are answer is correct.

  2. Check that the predicted direction for the reaction makes sense.

    If Q is too large compared to Kc (90 compared to 1.0 × 10-5), then the numerator in the mass-action expression is too big compared to the denominator.
    Therefore, in order to reach equilibrium, the concentration values in the numerator must be reduced while the concentration values in the denominator must increase.
    This means the reaction must be proceeding from right to left, consuming "products" and producing "reactants".
    That is, the reaction is proceeding in the reverse direction.

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

Q > Kc so the reaction will proceed in the reverse direction until equilibrium is achieved.

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Worked Example: Q < K

Question:

For the reaction: H2(g) + I2(g) ⇋ 2HI(g)
the equilibrium constant, K, is 1.59 × 102 at 500 K.
Predict the direction the reaction will move in if the reactants and products have the following concentrations:
[H2] = 0.15 mol L-1
[I2] = 0.75 mol L-1
[HI] = 1.75 mol L-1

Solution:

(Based on the StoPGoPS method for problem solving.)

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

Predict the direction the reaction will proceed in order to attain equilibrium.

PAUSE PAUSE to Prepare a Game Plan
 
  1. What data have you been given?

    Reaction: H2(g) + I2(g) ⇋ 2HI(g)
    Initial Concentrations:
    (mol L-1)    		 0.15   0.75   1.75
    Kc = 1.59 × 102 at 500 K

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

    (i) Q =   [HIg]2  
    [H2(g)][I2(g)]

    (ii)
    Q < K reaction proceeds in the forward direction
    Q = K system is at equilibrium,
    rate of forward reaction = rate of reverse reaction
    Q > K reaction proceeds in the reverse direction

GO GO with the Game Plan
  (i) Substitute the values for the concentration of each species into the mass-action expression, Q:
Q =   [HI]2  
[H2][I2]
Q =   [1.75]2  
[0.15][0.75]
Q = 27.22

(ii) Compare the value of the mass-action expression, Q, with the given value for the equilibrium constant, Kc.

Q = 27.22
Kc = 1.59 × 102

Q < K therefore reaction proceeds in the forward direction.

PAUSE PAUSE to Ponder Plausibility
  Is your answer plausible?
  1. Check your orders of magnitude
    Simplifiy the values for concentration by rounding to the nearest 10 in order to quickly decide if you have the correct order of magnitude:
    mass-action expression:Q =   [HI]2  
    [H2][I2]    		  
    Calculation we did above:Q =   [1.75]2  
    [0.15][0.75]    		 = 27.22
    Calculation using approximations:Q ≈   [1]2  
    [10-1][1]    		 =   1  
    10-1    		 = 101 = 10

    Since our approximation for Q (10) is of the same order of magnitude as our calculated value for Q (27.22) we are confident are answer is correct.

  2. Check that the predicted direction for the reaction makes sense.

    If Q is too small compared to Kc (27.22 compared to 1.59 × 102), then the numerator in the mass-action expression is too small compared to the denominator.

    Therefore, in order to reach equilibrium, the concentration values in the numerator must be increased while the concentration values in the denominator must decrease.

    This means the reaction must be proceeding from left to right, consuming "reactants" and producing "products".

    That is, the reaction is proceeding in the forward direction.

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

Q < Kc so the reaction will proceed in the foward direction until equilibrium is achieved.

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Worked Example: Q = K

Question:

For the reaction: N2O4(g) ⇋ 2NO2(g)
the equilibrium constant, Kc, is 5.0 × 10-1 at 100°C.
Predict the direction the reaction will move in if the concentration of N2O4 is 0.02 mol L-1 and the concentration of NO2 is 0.10 mol L-1.

Solution:

(Based on the StoPGoPS method for problem solving.)

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

Predict the direction the reaction will proceed in order to attain equilibrium.

PAUSE PAUSE to Prepare a Game Plan
 

  1. What data have you been given?

    Reaction: N2O4(g) ⇋ 2NO2(g)
    Initial Concentrations:
    (mol L-1)    		 0.02   0.10
    Kc = 5.0 × 10-1 at 100°C

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

    (i) Q =   [NO2g]2  
    [N2O4(g)]

    (ii)
    Q < K reaction proceeds in the forward direction
    Q = K system is at equilibrium,
    rate of forward reaction = rate of reverse reaction
    Q > K reaction proceeds in the reverse direction
GO GO with the Game Plan
  (i) Substitute the values for the concentration of each species into the mass-action expression, Q:

Q =   [NO2]2  
[N2O4]
Q =   [0.10]2  
[0.02]
Q = 0.50

(ii) Compare the value of the mass-action expression, Q, with the given value for the equilibrium constant, Kc.

Q = 0.5 = 5.0 × 10-1
Kc = 5.0 × 10-1

Q = K therefore the reaction is at equilibrium, that is, the rate of the forward reaction is equal to the rate of the reverse reaction so neither direction is favoured.

PAUSE PAUSE to Ponder Plausibility
  Is your answer plausible?

  1. Check your orders of magnitude
    Simplifiy the values for concentration by rounding to the nearest 10 in order to quickly decide if you have the correct order of magnitude:
    mass-action expression:Q =   [NO2]2  
    [N2O4]    		  
    Calculation we did above:Q =   [0.10]2  
    [0.02]    		 = 0.50
    Calculation using approximations:Q ≈   [10-1]2  
    [10-1]    		 =   10-2  
    10-1    		 = 10-1 = 0.1

    Since our approximation for Q (10-1) is of the same order of magnitude as our calculated value for Q (5.0 × 10-1) we are confident are answer is correct.

  2. Check that the predicted direction for the reaction makes sense.

    Q = Kc is an expression of the equilibrium law, that is, the rate of the forward reaction is the same as the reverse reaction, no changes are required to the relative sizes of the numerator and denominator in the mass-action expression.

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

Q = Kc so the reaction is at equilibrium and neither the forward nor the reverse direction is favoured.

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