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

Byju's Answer

Standard XII

Chemistry

Introduction

When 36.8g ...

Question

When 36.8g $N_{2} O_{4} (g)$ is introduced into a 1.0-litre flask at $27^{\circ} C$ . The following equilibrium reaction occurs: $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g); K_{p} = 0.1642 a t m$ . Calculate $K_{c}$ of the equilibrium reaction.

6.667 \times 10^{- 3} m o l L^{- 1}

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

6.999 \times 10^{- 3} m o l L^{- 1}

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

7.342 \times 10^{- 3} m o l L^{- 1}

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

7.893 \times 10^{- 3} m o l L^{- 1}

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

Open in App

Solution

The correct option is A $6.667 \times 10^{- 3} m o l L^{- 1}$
As we know,
$K_{p} = K_{c} (R T)^{Δ n}$
$K_{c} = K_{p} (R T)^{- Δ n} = 0.1642 (0.083 \times 300)^{-} = 6.667 \times 10^{- 3} m o l / L .$

Suggest Corrections

Similar questions

Q. When

36.6 g N_{2} O_{4} (g)

is introduced into a

1.0 - l i t r e

flask at

27^{\circ} C

. The following equilibrium reaction occurs:

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g); K_{p} = 0.1642 a t m

.
(a) Calculate

K_{c}

of the equilibrium reaction?
(b) What are the number of moles of

N_{2} O_{4}

and

N O_{2}

at equilibrium?
(c) What is the total gas pressure in the flask at equilibrium?
(d) What is the percent dissociation of

N_{2} O_{4}

Q. When 36.8g

N_{2} O_{4}

is introduced into a 1 litre flask at 300 K. The following reaction occurs:

N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g)

K_{P} = 0.1642

atm.
Calculate the

K_{C}

of the equilibrium reaction.

P C l_{5}, P C l_{3} a n d C l_{2}

are at equilibrium at

500

K in a closed container and their concentrations are

0.8 \times 10^{- 3} m o l L^{- 1}, 1.2 \times 10^{- 3} m o l L^{- 1}

and

1.2 \times 10^{- 3} m o l L^{- 1}

respectively.
The value of

K_{c}

for the reaction:

P C l_{5 (g)} ⇌ P C l_{3 (g)} + C l_{2 (g)}

will be:

For the reaction, $N_{2} O_{5} (g) \to N O_{2} (g) + \frac{1}{2} O_{2} (g)$ The value of rate of disappearance of $N_{2} O_{5}$ is given as $6.25 \times 10^{- 3} m o l L^{- 1} s^{- 1} .$ The rate of formation of $N O_{2}$ and $O_{2}$ is given respectively as:

Q. A mixture of

H_{2}, N_{2}

and

N H_{3}

with molar concentration

5 \times 10^{- 3} m o l L^{- 1}, 4 \times 10^{- 3} m o l L^{- 1} a n d 2 \times 10^{- 3} m o l L^{- 1}

respectively was prepared and heated to 500K.The value of

K_{c}

for the reaction

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

at this given temperature is 60.Predict whether ammonia tends to form or decompose at this stage of concentration

Join BYJU'S Learning Program

When 36.8g N2O4(g) is introduced into a 1.0-litre flask at 27∘C. The following equilibrium reaction occurs: N2O4(g)⇌2NO2(g);Kp=0.1642atm. Calculate Kc of the equilibrium reaction.

The correct option is A 6.667×10−3molL−1As we know,Kp=Kc(RT)ΔnKc=Kp(RT)−Δn=0.1642(0.083×300)−=6.667×10−3mol/L.

When 36.8g $N_{2} O_{4} (g)$ is introduced into a 1.0-litre flask at $27^{\circ} C$ . The following equilibrium reaction occurs: $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g); K_{p} = 0.1642 a t m$ . Calculate $K_{c}$ of the equilibrium reaction.

The correct option is A $6.667 \times 10^{- 3} m o l L^{- 1}$
As we know,
$K_{p} = K_{c} (R T)^{Δ n}$
$K_{c} = K_{p} (R T)^{- Δ n} = 0.1642 (0.083 \times 300)^{-} = 6.667 \times 10^{- 3} m o l / L .$