At a certain temperature, the equilibrium constant Kc is 0.25 for the reaction, A2(g)+B2(g)⇌C2(g)+D2(g)
If we take 1 mole of each of the four gases in a 10 litre container, what would be the equilibrium concentration of A2(g) ?
A
0.331 M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.033 M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.133 M
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
1.33 M
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C 0.133 M Concentration=MolesVolume
Volume= 10 L
Each species has 1 mole.
So concentration of each species : Concentration=110M=0.1M QC=[C2][D2][A2][B2]=0.1×0.10.1×0.1=1
Given that Kc=0.25
∵Qc>Kc so reaction will proceed in backward direction: A2(g)+B2(g)⇌C2(g)+D2(g)Conc.eqm1+x101+x101−x101−x10 0.25=(1−x10)2(1+x10)20.5=(1−x)(1+x)1.5x=0.5⇒x=0.333
[A2(g)]=1+x10=1.33310=0.133
Theory:
Application of Reaction Quotient(Q) :
Case 1 ReactionQuotient(Q)>EquilibriumconstantKeq;
For a given reaction R⟶P
When Q>Keq the product concentration [P] is very high as compared to concentration of reactants [R]. Since concentration of products is more so there will be more number of collisions among product molecules and product starts converting back to reactants until Q becomes equal to Keq i.e. If Q>Keq then it favours the backward reaction.
Case 2 ReactionQuotient(Q)<EquilibriumconstantKeq:
For a given reaction R⟶P
When Q<Keq the reactant concentration [R] is very high as compared to concentration of products [P]. Since concentration of reactants is more so there will be more number of collisions among reactant molecules and reactant starts converting back to products until Q becomes equal to Keq i.e. If Q<Keq then it favours the forward reaction.
Note :- Equilibrium is achieved when the rate of backward reaction is equal to the rate of forward reaction and not when the concentration of reactants and products becomes equal.