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N2o4 is 20% dissociated at 27 degree Celsius and total 1 ATM pressure calculate the percentage dissociation at 0.2 ATM and 27 degree Celsius?
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N2o4 is 20% dissociated at 27 degree Celsius and total 1 ATM pressure ...
Percentage Dissociation of N2O4 at 1 ATM and 27°C

To calculate the percentage dissociation of N2O4 at 0.2 ATM and 27°C, we first need to understand the concept of dissociation and how it relates to pressure and temperature.

Dissociation of N2O4
N2O4 (dinitrogen tetroxide) is a covalent compound that can dissociate into two molecules of NO2 (nitrogen dioxide) according to the following equation:

N2O4 ⇌ 2NO2

The extent of dissociation of N2O4 depends on the pressure and temperature conditions. At higher pressures and temperatures, the dissociation is more significant.

Le Chatelier's Principle
Le Chatelier's Principle states that when a system at equilibrium is subjected to a change in conditions, it will adjust to minimize the effect of that change. In the case of the dissociation of N2O4, increasing the pressure or temperature will shift the equilibrium towards the formation of more NO2.

Percentage Dissociation at 1 ATM and 27°C
Given that N2O4 is 20% dissociated at 27°C and 1 ATM pressure, we can assume that the equilibrium constant (Kp) for the dissociation reaction is 0.20. This means that at equilibrium, the concentration of NO2 is 0.20 times the concentration of N2O4.

If we let x represent the concentration of N2O4 that dissociates, the equilibrium concentrations can be expressed as follows:

[N2O4] = (1 - x) moles/L
[NO2] = 2x moles/L

Using the equilibrium constant expression, Kp = [NO2]^2 / [N2O4], we can substitute the equilibrium concentrations:

0.20 = (2x)^2 / (1 - x)

Simplifying the equation gives:

0.20 = 4x^2 / (1 - x)

Rearranging the equation:

0.20(1 - x) = 4x^2

0.20 - 0.20x = 4x^2

4x^2 + 0.20x - 0.20 = 0

Solving this quadratic equation will give us the value of x, which represents the concentration of N2O4 that dissociates.

Percentage Dissociation at 0.2 ATM and 27°C
To calculate the percentage dissociation at 0.2 ATM and 27°C, we need to consider the change in pressure. According to Le Chatelier's Principle, decreasing the pressure will shift the equilibrium towards the side with fewer moles of gas, which in this case is the N2O4 side.

Since the dissociation of N2O4 results in the formation of two moles of NO2 gas, decreasing the pressure will favor the reverse reaction, leading to a decrease in the percentage dissociation.

Therefore, the percentage dissociation of N2O4 at 0.2 ATM and 27°C will be lower than 20%. The exact value can be calculated using the same equilibrium constant expression and the new equilibrium concentrations.

However, without the value of the equilibrium constant at the new pressure, it is
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N2o4 is 20% dissociated at 27 degree Celsius and total 1 ATM pressure calculate the percentage dissociation at 0.2 ATM and 27 degree Celsius?
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