Question

what will be the total pressure of the gas mixture ...............plzzz help me solve this question.​

Answer 1

3) ɪꜱ ᴛʜᴇ ᴀɴꜱᴡᴇʀ...............

Answer 2

Answer:

The correct answer is P = $7 K_{P} ^{\frac{1}{2} }$

Explanation:

From the above question,

They have given :

QUESTION :

What will be the total pressure of the gaseous mixture at equilibrium if A$B_{2}$

is 20% dissociated according to the following reaction,  A$B_{2}$ (g) ↔ A(g) + 2B(g) ( $K_{p}$ represents the equilibrium constant of the reaction)

The total pressure of the gaseous mixture at equilibrium will be equal to the sum of the partial pressures of A and B.

Since the reaction is 20% dissociated, the partial pressure of A and B are 0.2Kp and 0.4Kp respectively.

Therefore, the total pressure of the gaseous mixture at equilibrium will be 0.2Kp + 0.4Kp = 0.6Kp.

Therefore, the total pressure of the gaseous mixture at equilibrium can be calculated as:

Total Pressure = Pₐ + 2Pₐ + 2Pᵦ

Where Pₐ is the partial pressure of component A and Pᵦ is the partial pressure of component B.

Since the reaction is 20% dissociated, the partial pressure of component A is 0.2 Kp and the partial pressure of component B is 0.4 Kp.

Therefore, the total pressure of the gaseous mixture at equilibrium is equal to 0.2Kp + 2(0.2Kp) + 2(0.4Kp) = 1.2Kp.

Hence,

The correct answer is P = $7 K_{P} ^{\frac{1}{2} }$

For more such related questions : https://brainly.in/question/38096509

#SPJ3