Question

5
1 mole of 'A', 1.5 mole of 'B' and 2 mole of 'C' are
taken in a vessel of volume one litre. At equilibrium
concentration of C is 0.5 mole/L. Equilibrium
constant for the reaction, Ag + Bg = Cg is
(1) 0.66
(2) 0.066
(3) 66
(4) 6.6

Answer 1

Answer:

A+B⇄C

  1         1.5         2

1-x        1.5-x     2+x=0.5

                         x=0.5-2

                         x=-1.5

                      2+x=2-1.5=0.5

$K_{c} =\frac{0.5}{2.5*3}$

               =$\frac{0.5}{7.5}$

               =0.0666

so option 2 is correct

Answer 2

The value of the equilibrium constant is 0.066

Explanation:

Initial concentration of $A=\frac{\text {initial moles of A}}{volume}=\frac{1mole}{1L}=1M$  

Initial concentration of $B=\frac{\text {initial moles of B}}{volume}=\frac{1.5mole}{1L}=1.5M$  

Initial concentration of $C=\frac{\text {initial moles of C}}{volume}=\frac{2mole}{1L}=2M$  

equilibrium concentration of C = 0.5 M

The given balanced equilibrium reaction is,

                            $A(g)+B(g)\rightleftharpoons C(g)$

Initial conc.         1 M      1.5 M         2 M

At eqm. conc.    (1+x) M   (1.5+x) M   (2-x) M

As concentration of C is decreasing, the reaction is taking place in backward direction.

The expression for equilibrium constant for this reaction will be,

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

$K_c=\frac{(2-x)}{(1+x)(1.5+x)}$

we are given : 2 - x = 0.5

x = 2 - 0.5 = 1.5

Now put all the value of x in this expression, we get :

$K_c=\frac{(2-1.5}{(1.5+1.5)(1+1.5)}$

$K_c=\frac{0.5}{(3.0)\times (2.5)}=0.066$

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