This is my question Into a 500mL container a chemist introduces 2.0 x 10^-2 mol of N2(g) and 2.0 x10^-2 mol of O2(g). The container is heated to 900degrees celcius and the following equilibrium is achieved.

Question

N2(g) + 2O2----N2O4(g)
calculate the intial concentrations in mol/L for each of the following three gases.
N2 would be 0.020/500mL x2 0.040mol/L
O2 would be 0.020/500ml x 2 0.040mol/L
N2O4 intial would be zero?
It then asks as equilibrium is reached 5.9 x 10^-3 mol of N2 is consumed. It change in concentration is therefore
0.0059-0.040= -0.0341
it then asks the change in O2 and N2O4?
This is where I get mixed up in the balanced equation we have 2O2 the intial was 0.040 would I multiply that by two since I have two moles to come up with a change in concentration of -0.068.
than N2O4 would be .1021?
Or am I all backwards and mixed up?
any help you can offer I would apprecitate your time.

Answer 1

I think you are using some of the numbers incorrectly.
.........N2 + O2 ==> N2O4
I......0.04..0.04.....0
C.......-x...-2x......+x
E...0.04-x...0.04-2x...+x

The problem states that the N2 consumed (that's x) is 0.0059 so the new (N2)is 0.04-0.0059 = ?
New (O2) is 0.04-[2*0.0059] = ?
New (N2O4) is 0+0.0059

Or to stay with your format, the change in concn N2 is -0.0059, the change for O2 is -2*0.0059 and the change in N2O4 is +0.0059

Answer 2

did you get them right?

Answer 3

in your second question, you have subtracted two terms with different units(mol/L - mol).

Thats y i asked