Le Chateliers principle tells us what will happen when equilibrium is disturbed qualitatively. This numerical problem is designed to explain the same principle quantitatively.This problem is large and its very purpose is to clear doubts regarding chemical equilibrium.
Consider a elementary reaction taking place in a closed vessel of volume We start with pure A with initial concentration of 100 .Reaction temperature and pressure are 300 K and 1 atm respectively. Other details are shown below.
We start with a simple question.
To attack this problem, we write expression for equilibrium concentration .
To solve for , we need to express in terms of .This can be done as follow
If this reaction would be irreversible,we would have 0 mol of A and 100 mol of B.
Remember Van't Hoff equation! That will give equilibrium constant at a particular temperature .For this, we need to solve following differential equation.
And integrating gives
We will use this relation to find at T=273K.
This will give .
From equilibrium constant formula
We see that is increased when temperature is lowered.
Similarly one can calculate at T=323 K
We notice that is increased when temperature is lowered.
These results prove Le chatelier principle :A exothermic reaction is favoured at low temperature as we get equilibrium concentration of B increases from 66.67 to 74.83 when temperature is lowered from 300K to 273K.
We know at T=300 K,
If we can relate concentration of A with time, problem can be easily solved by finding t for this .
For this we proceed as follow.
Material balance for A:
Rate of accumulation of A=input-output+generation-consumption
Since reaction is taking place in closed vessel,both input=0 and output=0
Rate of accumulation of A=
Putting these terms in matrial balance equation gives
Writing in terms of
Substituting values of , and
Above equation can be solved using method of integration factor.
where N is constant of integration.Its value can be found out by initial condition.
Now, we have relation between and t.
To find time required for equilibrium,put
This will give
This is not good.We did such lengthy calculation and got infinity as result.This may predict that equilibrium cannot be achieved in finite time.But if we find time required for to become 33.34, it will be only .This strange behavior can be explained by nature of function which become asymptote to the line .For practical purpose,time required for is enough.
We know a catalyst increases rate of reaction that too without actually get consumed.But here question is will it change equilibrium concentration of A and B.Answer is simply no.It increases both rate constant to the same extent so that ratio of rate constant that is is same as in non-catalytic reaction.For proof see How does catalyst work
For finding time required for equilibrium,we proceed as follow:
We have increased rate constant for both forward and backward reaction in catalytic reaction as
Now writing material balance for A,
Substituting values of , and
Try to solve this equation by integration factor method as we did above to get the final equation as
Putting and solving for t, we will get same result as we got above in case of non-catalytic reaction That is .This time we know how to avoid this problem.We proceed to find out time for We get
We see that time required for concentration to reach 33.34 is half of what we get in case of non catalytic reaction.
For non catalytic equation,we had
In general,One can easily show that
Time required will be less in catalytic reaction than non-catalytic reaction for same concentration.Here is the graph of concentration of A in two cases.
Since one mole of A gives one mole of B, there is no increase or decrease of pressure during reaction.Hence changing pressure at equilibrium will not change equilibrium concentration of A and B.
At equilibrium, we have 33.33 mol of A and 66.67 mol of B.Now we are adding another 50 mol of A.How this will disturb the equilibrium.Adding this extra A at previous equilibrium will result in same equilibrium as if it will be if we start with 100+50=150 mol of A.Another way to think is that 50 mol of A will give 33.33 mol B and 16.667 mol A will remain unreacted.So on total we will have 66.67+33.33=100 mol B and 33.33+16.6667=50 mol A.Hence we will have more B at equilibrium.
You can point out that percentage of B in equilibrium mixture is same in both cases(66.67%).But this is not in general.