N2O4 --->2NO2

a 1 liter flask is charged with .4 mols of N2O4. at equilibrium at 373 k, .0055 mols of N2O4 remain. what is the Kc for this reaction?

please explain to me how to get the rate law for this and solving

N2O4 ==> 2NO2

Set up an ICE chart.
initial:
N2O4 = 0.4
NO2 = 0

change:
N2O4 = -x
NO2 = +2x

equilibrium: (Just add initial to change to get equilibrium.)
N2O4 = 0.4 - x = 0.4-x
NO2 = 0 + 2x = 2x

The problem gives us that the equilibrium for N2O4 = 0.0055.
0.4- x = 0.0055
x = 0.4-0.0055 = ??

Now go to NO2. The change must be 2x and the total must be
NO2 = 2x

Now substitute those numbers into Kc expression.
Kc = (NO2)2/(N2O4)

To determine the equilibrium constant, Kc, for the given reaction, we need to write the expression for the equilibrium constant and plug in the given equilibrium concentrations.

The balanced equation is:
N2O4 ---> 2NO2

The expression for the equilibrium constant, Kc, is:
Kc = [NO2]^2 / [N2O4]

Given:
Initial concentration of N2O4 = 0.4 mol/L
Equilibrium concentration of N2O4 = 0.0055 mol/L

Let x be the amount of N2O4 that reacts. Since 2 moles of NO2 are formed for every 1 mole of N2O4 that reacts, the concentration of NO2 will be 2x at equilibrium.

Using the given information, we can calculate the equilibrium concentration of NO2:
Initial concentration of NO2 = 0 mol/L
Equilibrium concentration of NO2 = 2x mol/L

Since we know the equilibrium concentration of N2O4 is 0.0055 mol/L, we can substitute the values into the expression for Kc:

Kc = ([NO2]^2) / [N2O4]
Kc = [(2x)^2] / (0.0055 - x)

Now we need to solve for x. At equilibrium, the concentration of N2O4 is 0.0055 mol/L, so the amount that reacted is equal to the initial amount minus the equilibrium amount:

Amount of N2O4 that reacted = (0.4 - 0.0055) mol

Now, we can set up an equation to solve for x:

Amount of N2O4 that reacted = 0.4 - 0.0055 = x

Now, we can plug in this value for x into the expression for Kc:

Kc = [(2x)^2] / (0.0055 - x)

Kc = [(2 * (0.4 - 0.0055))^2] / (0.0055 - (0.4 - 0.0055))

Now, we can solve this equation to find the value of Kc.

To determine the equilibrium constant, Kc, for this reaction, you need to use the given information about the initial and equilibrium concentrations of N2O4.

Here's how you can get the rate law and solve for Kc:

Step 1: Write the balanced chemical equation:
N2O4 → 2NO2

Step 2: Determine the stoichiometry:
Since N2O4 reacts to form 2NO2, the molar concentration of NO2 at equilibrium will be twice that of N2O4.

Step 3: Set up an ICE table:

Initial Change Equilibrium
N2O4: 0.4 mol -x 0.4 - x mol
NO2: 0 mol +2x 2x mol

Step 4: Use the given information to find the equilibrium concentrations:
Given: 0.0055 mol N2O4 remains at equilibrium.

Since the stoichiometry tells us that 1 mol of N2O4 reacts to form 2 mol of NO2, we can say that the equilibrium concentration of NO2 is 2 * 0.0055 = 0.011 mol.

Step 5: Plug the equilibrium concentrations into the expression for Kc:
Kc = [NO2]^2 / [N2O4]

Kc = (0.011 mol)^2 / (0.4 - 0.0055 mol)
Kc = 0.000121 mol^2 / 0.3945 mol
Kc ≈ 0.000307

Therefore, the equilibrium constant (Kc) for this reaction is approximately 0.000307.

Please note that when performing calculations like this, it's important to double-check your units and ensure that they are consistent throughout the problem.

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