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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, H_{2(g)}, and iodine gas, I_{2(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 

hydrogen iodide gas 
⇋ 
hydrogen gas 
+ 
iodine gas 
concentration/ mol L^{1} 
3.531 × 10^{3} 

4.789 × 10^{4} 

4.789 × 10^{4} 
The system is at equilibrium, so, by writing the massaction expression^{2}and evaluating it using the concentrations of each species at equilibrium given above, we could obtain a value for the equilibrium constant, K_{c}, for this reaction at this temperature.
In order to write the massaction 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)} ⇋ ½H_{2(g)} + ½I_{2(g)}
 2HI_{(g)} ⇋ H_{2(g)} + I_{2(g)}
Which balanced equation we use determines the massaction expression we will write:

chemical equation 
massaction expression 
(1) 
HI_{(g)} ⇋ ½H_{2(g)} + ½I_{2(g)} 
Q_{(1)} = 
[H_{2(g)}]^{½}[I_{2(g)}]^{½} [HI_{(g)}] 

(2) 
2HI_{(g)} ⇋ H_{2(g)} + I_{2(g)} 
Q_{(2)} = 
[H_{2(g)}][I_{2(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 K_{c} since the system is at equilibrium):
(1) HI_{(g)} ⇋ ½H_{2(g)} + ½I_{2(g)}

massaction expression 
equilibrium constant 
Q_{(1)} = 
[H_{2(g)}]^{½}[I_{2(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)} ⇋ H_{2(g)} + I_{2(g)}

massaction expression 
equilibrium constant 
Q_{(2)} = 
[H_{2(g)}][I_{2(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} 
[H_{2(g)}][I_{2(g)}] [HI_{(g)}]^{2} 

= 
[H_{2(g)}]^{½}[I_{2(g)}]^{½} [HI_{(g)}] 
× 
[H_{2(g)}]^{½}[I_{2(g)}]^{½} [HI_{(g)}] 

= 
(  [H_{2(g)}]^{½}[I_{2(g)}]^{½} [HI_{(g)}]  )^{2} 

K_{(2)}  =  K_{(1)} × K_{(1)}  =  K_{(1)}^{2} 
0.0184 
= 
0.136 × 0.136 
= 
0.136^{2} = 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)} 
[H_{2(g)}]^{½}[I_{2(g)}]^{½} [HI_{(g)}] 

= 
√(  [H_{2(g)}][I_{2(g)}] [HI_{(g)}]^{2}  ) 

K_{(1)}  =  √K_{(2)} 
0.136 
= 
√0.0184 = 0.136 
The value of K_{c} 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 K_{c} and the relevant balanced chemical equation, you can derive a different value for K_{c} based on a different way to balance the chemical equation.
Worked Example
Question: Ammonia gas can be produced when nitrogen gas reacts with hydrogen gas according to the following balanced chemical equation:
N_{2(g)} + 3H_{2(g)} ⇋ 2NH_{3(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:
½N_{2(g)} + 1½H_{2(g)} ⇋ NH_{3(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


 What data have you been given?
(1) N_{2(g)} + 3H_{2(g)} ⇋ 2NH_{3(g)} K_{(1)} = 39 (at 400°C)
(2) ½N_{2(g)} + 1½H_{2(g)} ⇋ NH_{3(g)} K_{(2)} = ? (at 400°C)
 What is the relationship between what you have been given and what you need to find?
(1) 
N_{2(g)} + 3H_{2(g)} ⇋ 2NH_{3(g)} 
K_{(1)} = 
[NH_{3(g)}]^{2} [N_{2(g)}][H_{2(g)}]^{3} 

(2) 
½N_{2(g)} + 1½H_{2(g)} ⇋ NH_{3(g)} 
K_{(2)} = 
[NH_{3(g)}] [N_{2(g)}]^{½}[H_{2(g)}]^{3/2} 

K_{1} = K_{2}^{2}
K_{2} = √K_{1}

GO 
GO with the Game Plan 

Subsitute the value given for K_{(1)} into the equation:
K_{2} = √K_{1}
K_{2} = √39 = 6.24

PAUSE 
PAUSE to Ponder Plausibility 

Is your answer plausible?
Work backwards: is
K_{(1)} = K_{(2)}^{2} using our calculated value for K_{2} ?
K_{(1)} = 6.24^{2} = 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 :
½N_{2(g)} + 1½H_{2(g)} ⇋ NH_{3(g)}
K_{c} = 6.24
