A Problem On Equilibrium Constant

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 1m31 m^3 A(g)B(g)A_{(g)} \rightleftharpoons B_{(g)} We start with pure A with initial concentration of 100 mol/m3mol/m^3.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 KcK_c.

Kc=CbeCaeK_c=\frac{C_{be}}{C_{ae}}

Kc=kfkb=21=2K_c=\frac{k_f}{k_b}=\frac{2}{1}=2

CbeCae=2\frac{C_{be}}{C_{ae}}=2

To solve for CaeC_{ae}, we need to express CbeC_{be} in terms of CaeC_{ae}.This can be done as follow

Cbe=Ca0CaeC_{be}=C_{a0}-C_{ae}

Ca0CaeCae=2\frac{C_{a0}-C_{ae}}{C_{ae}}=2

Solving for CaeC_{ae}

Cae=33.33 mol/m3C_{ae}=33.33 \space mol/m^3

Cbe=10033.33=66.67 mol/m3C_{be}=100-33.33=66.67 \space mol/m^3

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.

d(lnKc)dT=ΔHRT2\frac{d(\ln K_c)}{dT}=\frac{\Delta H}{RT^2}

Separating variables,

d(lnKc)=ΔHRdTT2d(\ln K_c)=\frac{\Delta H}{R}\frac{dT}{T^2}

And integrating gives

lnK2K1=ΔHR(1T11T2)\ln \frac{K_2}{K_1}=\frac{\Delta H}{R}(\frac{1}{T_1}-\frac{1}{T_2})

We will use this relation to find KcK_c at T=273K.

ΔH=104\Delta H=-10^4

K1=2atT1=300KK_1=2\quad at \quad T_1=300 K

K2=?atT2=273KK_2=?\quad at \quad T_2=273 K

This will give K2=2.97K_2=2.97.

From equilibrium constant formula

Ca0CaeCae=2.97\frac{C_{a0}-C_{ae}}{C_{ae}}=2.97

Cae=25.17 mol/m3C_{ae}=25.17\space mol/m^3

And Cbe=10025.17=74.83 mol/m3C_{be}=100-25.17=74.83\space mol/m^3

We see that CbeC_{be} is increased when temperature is lowered.

Similarly one can calculate KcK_c at T=323 K

K323=1.50K_{323}=1.50

Cae=39.95 mol/m3C_{ae}=39.95\space mol/m^3

And Cbe=10039.95=60.05 mol/m3C_{be}=100-39.95=60.05\space mol/m^3

We notice that CbeC_{be} 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 mol/m3mol/m^3 to 74.83 mol/m3mol/m^3 when temperature is lowered from 300K to 273K.

We know at T=300 K, Cae=33.33 mol/m3C_{ae}=33.33\space mol/m^3

If we can relate concentration of A with time, problem can be easily solved by finding t for this CaeC_{ae}.

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=dCadt\frac{dC_a}{dt}

Generation=kbCbk_bC_b

Consumption=kfCak_fC_a

Putting these terms in matrial balance equation gives

dCadt=kbCbkfCa\frac{dC_a}{dt}=k_bC_b-k_fC_a

Writing CbC_b in terms of CaC_a

Cbe=Ca0CaeC_{be}=C_{a0}-C_{ae}

dCadt=kb(Ca0Cae)kfCa\frac{dC_a}{dt}=k_b(C_{a0}-C_{ae})-k_fC_a

dCadt=kbCa0(kf+kb)Ca\frac{dC_a}{dt}=k_bC_{a0}-(k_f+k_b)C_a

dCadt+(kf+kb)Ca=kbCa0\frac{dC_a}{dt}+(k_f+k_b)C_a=k_bC_{a0}

Substituting values of kf=2s1k_f=2 s^{-1},kb=2s1k_b=2 s^{-1} and Ca0=100 mol/m3C_{a0}=100 \space mol/m^3

dCadt+3Ca=100\frac{dC_a}{dt}+3C_a=100

Above equation can be solved using method of integration factor.

Ca=1003+Ne3tC_a=\frac{100}{3}+Ne^{3t}

where N is constant of integration.Its value can be found out by initial condition.

att=0,Ca=Ca0=100 mol/m3at \quad t=0,C_a=C_{a0}=100 \space mol/m^3

Which gives N=2003N=\frac{200}{3}

Ca=1003+200e3t3C_a=\frac{100}{3}+\frac{200e^{-3t}}{3}

Now, we have relation between CaC_a and t.

To find time required for equilibrium,put Cae=1003C_{ae}=\frac{100}{3}

1003=1003(1+2e3t)\frac{100}{3}=\frac{100}{3}(1+2e^{-3t})

This will give

e3t=0e^{-3t}=0

That is t=t=\infty

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 CaC_a to become 33.34, it will be only t=3.07st=3.07 s.This strange behavior can be explained by nature of function e3te^{-3t} which become asymptote to the line C=100/3C=100/3.For practical purpose,time required for Cae=33.34C_{ae}=33.34 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 KcK_c 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

kfc=4s1 and kbc=2s1k_{fc}=4 s^{-1} \space and \space k_{bc}=2 s^{-1}

Now writing material balance for A,

dCadt+(kfc+kbc)Ca=kbCa0\frac{dC_a}{dt}+(k_{fc}+k_{bc})C_a=k_bC_{a0}

Substituting values of kfc=2s1k_{fc}=2 s^{-1},kbc=2s1k_{bc}=2 s^{-1} and Ca0=100 mol/m3C_{a0}=100 \space mol/m^3

dCadt+6Ca=200\frac{dC_a}{dt}+6C_a=200

Try to solve this equation by integration factor method as we did above to get the final equation as

Ca=1003+200e6t3C_a=\frac{100}{3}+\frac{200e^{-6t}}{3}

Putting Cae=1003C_{ae}=\frac{100}{3} and solving for t, we will get same result as we got above in case of non-catalytic reaction That is t=t=\infty.This time we know how to avoid this problem.We proceed to find out time for Ca=33.34C_{a}=33.34 We get t=1.56st=1.56 s

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

Ca=1003+200e3t3C_a=\frac{100}{3}+\frac{200e^{-3t}}{3}

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.

Note by Aamir Faisal Ansari
3 years, 11 months ago

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