Question

when 3 mole of A and 1 mole of B are mixed in a ,the following reaction takes place

Answer 1

Ans should be a....please check calculation its comes out approx 0.1 after rounding off

Answer 2

Answer: 1) 0.12

Explanation:

The given balanced equilibrium reaction is,

$A(g)+B(g)\rightleftharpoons 2C(g)$

Moles of  A= 3 mole

Moles of  B  = 1 mole

Volume of solution = 1 L

Initial concentration of A = $\frac{moles}{\text {volume in liters}}=\frac{3}{1}=3M$

Initial concentration of B= $\frac{moles}{\text {volume in liters}}=\frac{1}{1}=1M$

                      $A(g)+B(g)\rightleftharpoons 2C(g)$

Initial conc.          3 M            1 M                    0

At eqm. conc.     (3-x) M      (1-x) M         (2x) M

Concentration of C at equilibrium = $\frac{moles}{\text {volume in liters}}=\frac{0.5}{1}=0.5M$

Given

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[C]^2}{[A][B]}$

given 2x = 0.5

x= 0.25

Now put all the given values in this expression, we get :

$K_c=\frac{(2x)^2}{(3-x)\times (1-x)}$  

x =  0.25

$K_c=\frac{[0.5]^2}{[3-0.25][1-0.25]}$

$K_c=0.12$

Thus equilibrium constant is 0.12