12. Take the equilibrium reaction X + Y ⇌ 2Z
The concentrations of X , Y and Z are 1, 1 and 2 moles per liter respectively. The equilibrium constant at the temperature is 5. In which way does the net reaction proceed?
From Quiz Concepts of Acids and Bases II
Answer:
Forward
Notice that it is not mentioned that the given concentrations are at equilibrium. In order to determine whether or not they are at equilibrium, we need to find the concentration/reaction quotient, which is the product of the concentrations of the products divided by that of the reactants, to the power of their respective reaction coefficients.
If the concentration quotient is the same as the value of the equilibrium constant, then the reaction is at equilibrium and so, the net reaction will not proceed anywhere.
Upon calculation, it is found out that the concentration quotient is just ([Z]^2/[X]*[Y]= 2^2/1*1=) 4. Since the value is less than the equilibrium constant, the concentration of products needs to increase, i.e. the net reaction shifts forward.
You can also calculate the final concentrations using the ICE*, or Initial-Change-Equilibrium method:-
Our initial concentrations are given in the question - the respective concentrations of X,Y and Z are 1, 1 and 2 molars respectively.
Let the concentration of moles of X or Y reacted be 'n'. Then the concentration moles of Z added would be 2n.
The equilibrium concentrations of X,Y and Z would be (1-n), (1-n) and (2+2n) respectively.
The equilibrium expression would be :-
Kc = (Concentration of Z)^2 / [ (1-n)*(1-n) ]
5 = (2+2n)^2 / [ (1-n)*(1-n)]
You will get a quadratic equation. Quadratic equations give two values for n, pick the value that is consistent. (Hint : If you pick the greater value of n, (1-n) would be negative, which isn't possible.)
You will get approximately n=0.06, and the equilibrium concentrations of X, Y and Z would be 0.94, 0.94 and 2.12 molars respectively.
*This isn't an official term, but simply a memory mnemonic to help you remember the steps.
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