1.00 mol of N2O4 and 1.00 mol of NO2 are placed in an 800 mL container. Calculate the initial concentrations of each gas.

N2O4= 1.25 mol/L
NO2=1.25 mol/L
When equilibrium is reached, the concentration of NO2 increases by 0.50mol/L. Calculate the equilibrium NO2 and N2O4
If I set up an ice table
N2O4 2NO2
initial 1.25 mol/L 1.25mol/L
change -0.25 0.50
equil 1.0 mol/L 1.75mol/L
I am sure this is not right because I should have double the 2NO2. If you could point out where I have gone astray I would appreciate it.

I don't see that you have gone astray. I don't see anything wrong.

It seems like you made a mistake in setting up the initial concentrations in your ICE table. Let's go through the calculations step by step and see where the error occurred.

Given:
Initial moles of N2O4 = 1.00 mol
Initial moles of NO2 = 1.00 mol
Volume of the container = 800 mL

To calculate the initial concentrations of each gas, we need to convert the volume from milliliters (mL) to liters (L).

Volume of the container = 800 mL = 800 mL / 1000 mL/L = 0.800 L

Now we can calculate the initial concentrations:

For N2O4:
Initial moles of N2O4 = 1.00 mol
Volume of the container = 0.800 L
Initial concentration of N2O4 (C1) = moles of N2O4 / volume = 1.00 mol / 0.800 L = 1.25 mol/L

For NO2:
Initial moles of NO2 = 1.00 mol
Volume of the container = 0.800 L
Initial concentration of NO2 (C2) = moles of NO2 / volume = 1.00 mol / 0.800 L = 1.25 mol/L

So, the correct initial concentrations of N2O4 and NO2 are both 1.25 mol/L.

Now, let's move on to the equilibrium calculations.

Given:
Change in concentration of NO2 = +0.50 mol/L

To find the equilibrium concentrations, we need to consider the stoichiometry of the reaction. From the equation N2O4 ⇌ 2NO2, we know that for every one mole of N2O4 that reacts, two moles of NO2 are formed.

Using the initial concentrations and the change in concentration, we can update the ICE table as follows:

N2O4 2NO2
Initial: 1.25 mol/L 1.25 mol/L
Change: -x +2x
Equilibrium: 1.25 - x 1.25 + 2x

The change in concentration for NO2 is given as +0.50 mol/L. Therefore, we can set up the equation:

0.50 mol/L = 1.25 + 2x

Now we can solve for x:

2x = 0.50 mol/L - 1.25 mol/L
2x = -0.75 mol/L
x = -0.375 mol/L

Since we cannot have a negative concentration, we discard this negative value and choose the positive root.

x = 0.375 mol/L

Now substitute this value back into the equilibrium concentrations:

Equilibrium concentration of N2O4 = 1.25 - x = 1.25 - 0.375 = 0.875 mol/L
Equilibrium concentration of NO2 = 1.25 + 2x = 1.25 + 2 * 0.375 = 1.25 + 0.75 = 2.00 mol/L

Therefore, the equilibrium concentrations are 0.875 mol/L of N2O4 and 2.00 mol/L of NO2.

I hope this clarifies the calculation process and helps you correct your mistake.