Question

A 1-litre vessel contains 2 moles each of gases A,
B, C and D at equilibrium. If one mole each of A
and B are removed, Kc for
A + BAC+D will be :-
(a) 4 (6) 1 (c) 1/4 (d) 2​

Answer 1

Answer:

option (a) ....................

Answer 2

$K_c$ for   $A+B\rightleftharpoons C+D$  will be 1

Explanation:

Initial moles of A = 2 mole

Volume of container = 1 L

equilibrium concentration of A = $\frac{moles}{volume}=\frac{2moles}{1L}=2M$  

equilibrium concentration of B = $\frac{moles}{volume}=\frac{2moles}{1L}=2M$  

equilibrium concentration of C= $\frac{moles}{volume}=\frac{2moles}{1L}=2M$  

equilibrium concentration of D= $\frac{moles}{volume}=\frac{2moles}{1L}=2M$  

The given balanced equilibrium reaction is,

                            $A+B\rightleftharpoons C+D$

At eqm. conc.        2M     2M                 2M      2M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[C][D]}{[A][B}}$

$K_c=\frac{3\times 3}{3\times 3}$

$K_c=1$

As the equilibrium will be reestablished, the equilibrium constant will remain same.

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