srry the Queation is : At 1000°C the pressure of iodine gas is found to be 0.112 atm whereas the expected pressure is 0.074 atm. the increased pressure is due to dissociation of I2 ? 2I . calculate Kp. also find out pressure at which I2 will be 90% dissociated at 1000°C.

To tackle this question, we need to analyze the dissociation of iodine gas (I2) into its atomic form (2I) and calculate the equilibrium constant (Kp) based on the given pressures. We’ll also determine the pressure at which 90% of I2 is dissociated at 1000°C. Let’s break this down step by step.

Understanding the Reaction

The dissociation of iodine gas can be represented by the following equilibrium reaction:

I2(g) ⇌ 2I(g)

When I2 dissociates, it produces two moles of iodine atoms (I) for every mole of I2 that breaks apart. This means that the number of moles of gas increases, which can lead to an increase in pressure.

Given Data

• Pressure of I2 at 1000°C: 0.112 atm
• Expected pressure of I2 (if no dissociation occurred): 0.074 atm

Calculating the Change in Pressure

To find out how much I2 has dissociated, we can use the difference between the actual pressure and the expected pressure. The increase in pressure indicates that some I2 has dissociated into I atoms.

Let’s denote:

• P_initial = 0.074 atm (pressure of I2 if no dissociation)
• P_actual = 0.112 atm (actual pressure)
• x = change in pressure due to dissociation

The change in pressure due to dissociation can be calculated as:

x = P_actual - P_initial = 0.112 atm - 0.074 atm = 0.038 atm

Establishing the Equilibrium Expression

At equilibrium, the pressure of I2 will be:

P_I2 = P_initial - x = 0.074 atm - 0.038 atm = 0.036 atm

The pressure of I (which is produced from the dissociation) will be:

P_I = 2x = 2 * 0.038 atm = 0.076 atm

Calculating Kp

The equilibrium constant Kp for the reaction can be expressed as:

Kp = (P_I)^2 / (P_I2)

Substituting the values we have:

Kp = (0.076 atm)^2 / (0.036 atm)

Kp = 0.005776 / 0.036 = 0.1604

Finding the Pressure for 90% Dissociation

To find the pressure at which 90% of I2 is dissociated, we can set up the following:

If 90% of I2 is dissociated, then:

• Amount of I2 remaining = 10% of initial = 0.1P_initial
• Amount of I produced = 90% of initial = 0.9P_initial (which will produce 1.8P_initial of I)

Let’s denote the initial pressure of I2 as P_initial. The total pressure at equilibrium will be:

P_total = P_I2 + P_I = 0.1P_initial + 1.8P_initial = 1.9P_initial

Using the equilibrium expression again:

Kp = (P_I)^2 / (P_I2)

Substituting the pressures:

Kp = (1.8P_initial)^2 / (0.1P_initial)

Kp = 3.24P_initial

Now, we can set this equal to the previously calculated Kp value:

0.1604 = 3.24P_initial

P_initial = 0.1604 / 3.24 = 0.0495 atm

Final Results

In summary, we have:

• Kp at 1000°C = 0.1604
• Pressure at which I2 will be 90% dissociated = 0.0495 atm

This analysis shows how the dissociation of a gas affects its equilibrium state and how to calculate the equilibrium constant and pressures involved. If you have any further questions or need clarification on any part of this process, feel free to ask!

Approved