You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Concentration Terms

Two gases A...

Question

Two gases $A$ and $B$ in the molar ratio 1 : 2 were admitted to any empty vessel and allowed to reach equilibrium at $400^{o}$ C and 8 atm pressure as : $A + 2 B$ $⇌$ $2 C$ . The mole fraction of $C$ at equilibrium is 0.16. Calculate:
(a) $K_{r}$ for the reaction,
(b) The pressure at which mole fraction of $A$ in equilibrium mixture is 0.16.

Open in App

Solution

Moles $A + 2 B ⇌ 2 C$ Total moles
at $t = 0$ $x$ $2 x$ -
at $E q b^{m}$ $x - a$ $2 x - 2 a$ $2 a$ $3 x - a$
Mole fraction of $C = \frac{2 a}{3 x - a} = 0.16$
$\Rightarrow 1.84 a = 0.48 x$ or $3 x - a = 12.5 a ⟶ (1)$
$K_{P} = \frac{[X_{c}]^{2}}{[X_{o}]^{2} [X_{A}]^{2}} [P T]^{Δ n g}$
$K_{P} = \frac{[0.16]^{2}}{[X_{B}]^{2} [x_{A}]} [P T]^{Δ n g}$
$\Rightarrow X_{A} + X_{B} = 1 - X_{C} = 0.84$
$\Rightarrow X_{A} + X_{B} = (x - a) + (2 x - 2 a) = 3 x - 3 a = (3 x - a) - 2 a$
$\Rightarrow 12.5 a - 2 a$ (from equation $1$ )
$\Rightarrow X_{A} + X_{B} = 10.5 a = 0.84$
$\Rightarrow a = 0.08$
$\Rightarrow x = \frac{1.08 \times a}{0.48} - \frac{1.84 \times 0.08}{0.48} = 0.306$
Thus, $X_{A} = \frac{x - a}{3 x - a} = \frac{0.306 - 0.08}{12.5 \times 0.08} = 0.227$
$X_{B} = 0.84 - 0.227 = 0.613$
$K_{P} = \frac{[0.16]^{2}}{[0.613]^{2} [0.227]} (8 a t m)^{- 1} = 0.0374$

Suggest Corrections

Similar questions

Q. Two gases

A

and

B

in the molar ratio

1 : 2

were admitted to any empty vessel and allowed to reach equilibrium at

400^{o} C

and

8 a t m

pressure as

A + 2 B ⇌ 2 C

. The mole fraction of

C

at equilibrium is

0.4

. Calculate

K_{P} (a t m^{- 1})

for the reaction.

Q. A reaction carried out by 1 mol of N

_{2}

and

3

mol of H

_{2}

shows at equilibrium the mole fraction of NH

_{3}

0.012

at 500

^{o}

C and

10

atm pressure. report the pressure at which mole % of

N H

_{3}

in equilibrium mixture is increased to

10.4

(in atm)

Q. The equilibrium constant for the reaction,

C O (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g)

986^{o} C

0.63

. A mixture of

1.0

mole of water vapor and

3.0

moles of

C O

is allowed to come to equilibrium. The equilibrium pressure is

2.0 a t m

(i) How many moles of

H_{2}

are present at equilibrium?
(ii) Calculate the partial pressure of gases in the equilibrium mixture.

Q. A reaction carried out by 1 mol of

N_{2}

and 3 mol of

H_{2}

at equilibrium. The mole fraction of

N H_{3}

as 0.012 at 500C and 10 atm pressure.What the pressure at which mole % of

N H_{3}

in equilibrium mixture is increased to 10.4?. (in atm)

Q. A reaction is carried out by 1 mole of

N_{2}

, 3 mole of

H_{2}

shows at equilibrium, the mole fraction of

N H_{3}

is 0.012 at 500

^{o} C

and 10 atm pressure.

K_{p}

is the equilibrium constant. The pressure (in atm) at which mole % of

N H_{3}

in equilibrium mixture increased to 10.4 is_________.
[Give your answer in closest integer]

Join BYJU'S Learning Program