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perin

k_c for A (g) ===== B (g) + C(g) is 0.45 at 473 K. 1 litre of container has 0.2

mole of A and 0.3 mole of B and 0.3 mole of C. calculate the new equilibrium

concentrations of A,B, C when volume of container is halved. ans = A = 0.52M
B=C= 0.48M

my doubt is that when volume is halved then we divide the moles of A,B,C by

0.5 to get their molarity and then proceed. but why do we divide ? we can do

by simply taking the moles given but that doesn't give the answer.

arpan1

A                           ------------       B           +    C

0.2            0.3   0.3
(O.2 - X)                                  (0.3 +X)                      (0.3 +X)

Total moles = 0.8 + X

THEN

0.45 = (0.3+ X) ² / (0.2 - X) (0.8 + X)

WE GET X = 0.05

Therefore at equilibrium
A = 0.15
B= 0.35
C=0.35

When volume is made half, acc. to le-chatelier principle , backward reaction is favoured.

B                 +           C   ---------------------------                  A

(0.35 - Y)             (0.35 - Y)                                 (0.15 + Y)

TOTAL MOLES = 0.85 - Y

1/0.45 = 0.5 (0.15 + Y ) ( 0.85 - Y )/ (0.35 - Y)²

Y = 0.28

THEN
A = 0.43
B= C = 0.07

RATE ME

perin

please sumbudy explain....my doubt.

it's a p bahadur question

solution is like this

when volume is halved reaction goes towards lesser no. of moles that is backwards

so B + C ======== A

0.3/0.5 - x 0.3/0.5 - x 0.2/0.5 + x

kc is given so put the values and solve for x and get the answer.

MY DOUBT why did we divide the moles of A B C by 0.5 when volume is halved.??? if we don't do that we don't get the answer

@ arpan :yur answer is incorrect

computer001

well c v have to devide it by 2 cuz:
PV=nRT
as nothing mentioned abt pressure v have to assume its constant...as otherwise it will affect eqbm....
so P,R,T are const
so as V->V/2
n->n/2
and here v have to c in terms of mole ratio
it like lets say v=10 L A=2L B=3L C=5L
as soo as V is halved it follows ob tht A=1L B=1.5 C=2.5
after this v find new eqbm conc

akshay.khare91

look the solution can be like this

first divide the moles of A and B and C divide it by volume V
to get concentration and write equation of Kc

then equate it with Kc to find V

then suppose that the moles of A are initially 1 and x moles
get dissociated then
moles. of A at equilibrium = 1-x
of B = x
of C = x

since change in volume will not change Kc therefore
divide these mole of V/2 and equate with Kc ..which will
gives x and then find moles...

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