Question

A 0.100 mol sample of N2O4 is placed in a 2.5L flask and allowed to reach equilibrium at 348K. The mole fraction of NO2 at equilibrium is 0.746. Determine Kp at 348K for the reaction N2O4 (g) <=> 2NO2 (g).

Answer 1

Step-by-step explanation:

Here, we want to get the equilibrium constant of the reaction

Firstly, we set up the initial, change and equilibrium chart as follows:

Since we have the equilibrium values, we can use the mole fraction to get the value of x

What the mole fraction is saying is: If we divided the number of moles of NO2 by the total, we have the mole fraction as 0.746. This could help us to get the value of x and by extension, get the value of the equilibrium constant

We can have it calculated as follows:

`\begin{gathered} (2x)/(2x+0.1-x)\text{ = 0.746} \\ \\ (2x)/(x+0.1)\text{ = 0.746} \\ \\ 2x\text{ = 0.746(x+0.1)} \\ 2x\text{ = 0.746x +0.0746} \\ 2x-0.746x\text{ = 0.0746} \\ 1.254x\text{ = 0.0746} \\ x\text{ = }(0.0746)/(1.254) \\ \\ x\text{ = 0.0595} \end{gathered}`

From here, we can get the equilibrium concentrations by supplying the value of x

For N2O4, we have 0.1 - 0.0595 = 0.0405 moles

For NO2, we have 2x = 2(0.0595) =