Calculation of KP for the Reaction A2 + B2 ⇌ 2AB

Given:

- 2 moles of gas A2
- 2 moles of gas B2
- Volume of the flask = 1 litre
- At equilibrium, 0.5 moles of A2 was obtained

To calculate KP for the reaction, we need to use the equation:

KP = (PAB)² / (PA² x PB²)

Where,

- PAB = the partial pressure of AB at equilibrium
- PA = the partial pressure of A2 at equilibrium
- PB = the partial pressure of B2 at equilibrium

Calculating Partial Pressures

Since we are given the number of moles of A2 and B2, we can use the ideal gas law to calculate their partial pressures:

PV = nRT

Where,

- P = pressure
- V = volume
- n = number of moles
- R = gas constant
- T = temperature

We can rearrange the equation to get:

P = nRT / V

Substituting the values given:

- For A2: PA = (0.5 mol x 8.31 J/mol.K x 298 K) / 1 L = 123.585 Pa
- For B2: PB = (2 mol x 8.31 J/mol.K x 298 K) / 1 L = 1970.16 Pa

Calculating Partial Pressure of AB

Since 0.5 moles of A2 reacts to form 0.5 moles of AB, the number of moles of AB at equilibrium is also 0.5.

Therefore, PAB = (0.5 mol x 8.31 J/mol.K x 298 K) / 1 L = 61.7925 Pa

Calculating KP

Substituting the values calculated:

KP = (PAB)² / (PA² x PB²)

= (61.7925 Pa)² / (123.585 Pa)² x (1970.16 Pa)²

= 1.24 x 10^-4 Pa^-2

Therefore, KP for the reaction A2 + B2 ⇌ 2AB is 1.24 x 10^-4 Pa^-2.

Explanation

The equilibrium constant KP is a measure of the extent to which a reaction will proceed towards the products at a given temperature and pressure. In this case, we are given the initial number of moles of A2 and B2 and the number of moles of A2 at equilibrium. Using the ideal gas law, we can calculate the partial pressures of A2 and B2 at equilibrium. Since the reaction involves gases, we use the partial pressures to calculate KP using the equation KP = (PAB)² / (PA² x PB²). The value of KP tells us that the reaction will not proceed much towards the products at the given temperature and pressure.

36 reasonkp=(1.5*2)^(2)/(0.5*0.5)=36