Question

40. The reaction, 2A(g) + B(g) = 3C(g) + D(g)
is begun with the concentrations of A and B both
at an initial value of 1.00 M. When equilibrium is
reached, the concentration of D is measured and
found to be 0.25 M. The value for the equilibrium
constant for this reaction is given by the
expression
[AIPMT (Mains)-2010]
(1) [(0.75)(0.25)] + [(1.00)?(1.00)]
(2) [(0.75)(0.25)] + [(0.50)2(0.75)]
(3) [(0.75)*(0.25)] = [(0.50)?(0.25)]
(4) [(0.75)3(0.25)] + [(0.75)2(0.25)]​

Answer 1

Answer:

Explanation:

2A   +   B   =   3C   +   d

1            1           0         0          at t=0

1-2x        1-x     3x         x           at equilibrium

concentration of d is found to be 0.25, hence x=0.25

0.5         0.75   0.75    0.25

k= (0.25)(0.75)³/(0.5)²(0.75)

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Answer 2

The value for the equilibrium  constant for this reaction is given by the

expression $\frac{(0.25)\times (0.75)}{(0.5)\times (0.25)}$

Explanation:

Initial concentration of $A=\frac{moles}{volume}=1.00M$  

Initial concentration of $B=\frac{moles}{volume}=1.00M$  

equilibrium concentration of D$=\frac{moles}{volume}=0.25M$

The given balanced equilibrium reaction is,

                         $2A(g)+B(g)\rightleftharpoons 3C(g)+D(g)$

Initial conc.                1 M    1 M           0     0  

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

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[A]^2\times {B}^1}{[C]^3\times [D]^1}$

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

we are given : x= 0.25 M

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

$K_c=\frac{(1-2\times 0.25)^2\times (1-0.25)}{(3\times 0.25)^3\times (0.25)}$

$K_c=\frac{(0.25)\times (0.75)}{(0.5)\times (0.25)}$

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