An equilibrium mixture of -160- CO - g - - -160-H-2 O - g - CO-2 - g - - H-2 - g - -160-present in a vessel of one litre capacity at - 815-C- -160-was found by analysis to contain 0-4 mole of CO- 0-3 mole of - H-2 O- 0-2 mole of - CO-2- -160-and 0-6 mole -160-of - H-2-a- Calculate - K-c-b- If it is derived to increase the concentration of -160-CO to 0-6 moles by adding - CO-2- -160-to the vessel- -160-how many moles must be added into equilibrium -160-mixture at a constant temperature in order to get -160-this change-1 and 75 moles1 and 0-075 mole2 and 7-5 moles1 and 0-75 moles

The correct option is D 1 and 0.75 moles K_p = [CO_2][H_2][CO][H_2O] K_p = 0.2 × 0.60.4×0.3 K_p = 1 Now we wish to find the number of moles of ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Derivation of Kp and Kc

An equilibriu...

Question

An equilibrium mixture of $C O (g) + H_{2} O (g) ⇋ C O_{2} (g) + H_{2} (g)$ , present in a vessel of one litre capacity at $815^{\circ} C$ was found by analysis to contain 0.4 mole of CO, 0.3 mole of $H_{2} O$ , 0.2 mole of $C O_{2}$ and 0.6 mole of $H_{2}$ .

(a) Calculate $K_{c}$ .
(b) If it is derived to increase the concentration of CO to 0.6 moles by adding $C O_{2}$ to the vessel, how many moles must be added into equilibrium mixture at a constant temperature in order to get this change?

2 and 7.5 moles

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 and 0.75 moles

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1 and 75 moles

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 and 0.075 mole

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D 1 and 0.75 moles
$K_{p} = \frac{[C O_{2}] [H_{2}]}{[C O] [H_{2} O]}$
$K_{p} = \frac{0.2 \times 0.6}{0.4 \times 0.3}$
$K_{p} = 1$
Now we wish to find the number of moles of $C O_{2}$ added so that the equilibrium shifts backward and finally there are 0.6 moles of $C O_{2}$ .
Since, $Δ n_{g} = 0$
$K_{p} = K_{c} = 1$
Suppose 'x' mole $C O_{2}$ is added. then,
$C O (g) + H_{2} O (g) \to C O_{2} (g) + H_{2} (g)$
Initially (0.4) (0.3) (0.2 + x) (0.6)

At equilibrium (0.4+0.2) (0.3+0.2) (0.2+x - 0.2) (0.6-0.2)
$K_{p} = \frac{0.6 \times 0.5}{x \times 0.4} = 1$
$⟹ x = 0.75$

Option B is correct.

Suggest Corrections

Similar questions

Q. An equilibrium mixture of

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

present in a vessel of one litre capacity at

815^{0}

C was found by analysis to contain 0.4 mol of

C O

, 0.3 mol of

H_{2} O

, 0.2 mol of

{C O}_{2}

and 0.6 mol of

H_{2}

a) Calculate

K_{C}

.
b) If it is derived to increase the concentration of

C O

to 0.6 mol by adding

{C O}_{2}

to the vessel, how many moles must be added into equilibrium mixture at a constant temperature in order to get this change?

Q. An equilibrium mixture

{S O}_{2} (g) + {N O}_{2} (g) ⇌ {S O}_{3} (g) + N O (g)

was found to contain

0.6

mole of

{S O}_{2}

0.2

mole of

{N O}_{2}

0.8

mole of

{S O}_{2}

and

0.3

mole of

N O

in a liter. vessel. How many moles of

{N O}_{2}

per it. will have to be added to the vessel in order to increase the

N O

concentration to

0.5

mole/liter.

Q. A two litre vessel contains 0.48 mole of

C O_{2}

, 0.48 mole of

H_{2}

, 0.96 mole of

C O

at equilibrium.

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

How many

g m

(only integer part) of

H_{2}

must be added to bring the concentration of

C O

0.6 M

Q. In a study of equilibrium,

H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)

1 mole of

H_{2}

and 3 mole of

I_{2}

gave rise at equilibrium to

x

mole of

H I

.
Addition of a further 2 mole of

H_{2}

gave an additional

x

mole of

H I

. What is

x

Q. A mixture of one mole of

C O_{2}

and one mole of

H_{2}

attains equilibrium at

250^{\circ} C

and 0.1 atm.

C O_{2} + H_{2} \to C O + H_{2} O

Calculate

K_{p}

, if the analysis of final reaction mixture shows 0.16 volume percent of CO :

Join BYJU'S Learning Program