Question

If 3 mole of a reactant A and one mole of a reactant b are mixed in a vessel of volume 1 litre.The reaction taking place is A+B=2C .If 1.5 mole of C is formed at equilibrium the valur of equilibrium constant is?

Answer 1

$\huge{\textbf{\underline{Equilibrium Condition:-}}}$

$\large{\textbf{\underline{Given :-}}}$

Volume of vessel = 1 L

mole of C formed at equilibrium = 1.5

$\huge{\textbf{\underline{Solution:-}}}$

• Refer to attachment.

$\textbf{ A + B \rightleftharpoons 2C }$

From attachment we conclude that the mole formed of C at equilibrium is 2x.

Then , mole of C formed at equilibrium is

2x = 1.5

x = 1.5/2

x = 0.75

Now we can use the formula for equilibrium constant:-

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

$\rightsquigarrow$$K_c = \left(\dfrac{2x^2}{(3-x)(1-x)}\right)$

• Concentration at equilibrium is in attachment.

$\rightsquigarrow$$K_c = \dfrac{( 2\times 0.75)^2}{(3-0.75)(1-0.75)}$

$\rightsquigarrow$$K_c = \dfrac{2.25}{2.25 \times 0.25}$

$\rightsquigarrow$$K_c = \dfrac{2.25}{0.5625}$

$\rightsquigarrow$$K_c = 4$

$\therefore$ $\huge{\textbf{\underline{ K_c = 4 }}}$

Answer 2

Answer:

Solution:-

Refer to attachment.

A+B⇌2C\textbf{$A + B \rightleftharpoons 2C$}A+B⇌2C

From attachment we conclude that the mole formed of C at equilibrium is 2x.

Then , mole of C formed at equilibrium is

2x = 1.5

x = 1.5/2

x = 0.75

Now we can use the formula for equilibrium constant:-

Kc=[C]2[A][B]\large{K_c = \dfrac{[C]^2}{[A] [B]}}K

c

=

[A][B]

[C]

2

⇝\rightsquigarrow⇝ Kc=(2x2(3−x)(1−x))K_c = \left(\dfrac{2x^2}{(3-x)(1-x)}\right)K

c

=(

(3−x)(1−x)

2x

2

)

Concentration at equilibrium is in attachment.

⇝\rightsquigarrow⇝ Kc=(2×0.75)2(3−0.75)(1−0.75)K_c = \dfrac{( 2\times 0.75)^2}{(3-0.75)(1-0.75)}K

c

=

(3−0.75)(1−0.75)

(2×0.75)

2

⇝\rightsquigarrow⇝ Kc=2.252.25×0.25K_c = \dfrac{2.25}{2.25 \times 0.25}K

c

=

2.25×0.25

2.25

⇝\rightsquigarrow⇝ Kc=2.250.5625K_c = \dfrac{2.25}{0.5625}K

c

=

0.5625

2.25

⇝\rightsquigarrow⇝ Kc=4K_c = 4K

c

=4

∴\therefore∴ Kc=4‾\huge{\textbf{\underline{$K_c = 4$}}}

K

c

=4