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Answer 1

Question :--

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

A(g) + B(g) -----> 2C(g)

the equilibrium mixture contains 0.5 mole of C what is the value of equilibrium constant for the reaction ..?

Answer :--

$given \: equilibrium \: reaction \: .$

A(g) + B(g) <==> 2C (g)

moles of A = 3 mole

moles of B = 1 mole

volume of the solution = 1 litre

initial concentration of A ( molarity)

= moles / volume

= 3/1

= 3M

initial concentration of B ( molarity)

= moles / volume

= 1/1

= 1M

A(g) + B(g) <==> 2C(g)

initial conc. 3M. 1M

equili. conc. (3-x)M. (1-x)M

concentration of C at the equilibrium (molarity)

= moles / volume

= 0.5 / 1

= 0.5 M

equilibrium constant = Kc

Kc = [c]^2 / [A][B]

here given 2x = 0.5

x = 0.5/2

x = 0.25

now put the values in the formula ..

$kc \: = \frac{(2x) {}^{2} }{(3 - x)(1 - x)}$

$kc \: = \frac{(0.5) {}^{2} }{(3 - 0.25)(1 - 0.25)}$

$kc \: = 0.12$

the equilibrium constant Kc = 0.12.

This is your answer..