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.
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.
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
did you get them right?
in your second question, you have subtracted two terms with different units(mol/L - mol).
Thats y i asked
Let's go step by step to calculate the initial concentrations and the changes in concentrations for each gas.
1. Initial Concentrations:
To calculate the initial concentrations of N2 and O2, you correctly used the given moles (0.020 mol) and the volume of the container (500 mL). However, it's important to convert the volume to liters before calculating the concentration. So the calculations should be as follows:
N2 initial concentration = (0.020 mol) / (0.500 L) = 0.040 mol/L
O2 initial concentration = (0.020 mol) / (0.500 L) = 0.040 mol/L
For N2O4, it is correct that the initial concentration is zero since it is not initially present and is formed during the reaction.
2. Changes in Concentrations:
Given that 5.9 x 10^-3 mol of N2 is consumed, the change in concentration of N2 can be calculated by subtracting the moles consumed from the initial concentration:
Change in [N2] = 0.040 mol/L - 5.9 x 10^-3 mol = 0.0341 mol/L (Note that the sign is negative because the concentration decreases)
Now, let's calculate the changes in concentration of O2 and N2O4.
Looking at the balanced equation: N2(g) + 2O2(g) -> N2O4(g)
We see that for every 1 mole of N2 consumed, 2 moles of O2 and 1 mole of N2O4 are formed.
Therefore, the change in [O2] would be:
Change in [O2] = 2 x (Change in [N2]) = 2 x (-0.0341 mol/L) = -0.0682 mol/L (Negative because concentration decreases)
Similarly, the change in [N2O4] would be:
Change in [N2O4] = 1 x (Change in [N2]) = 1 x (-0.0341 mol/L) = -0.0341 mol/L (Negative because concentration decreases)
So, to summarize:
- The change in [N2] is -0.0341 mol/L
- The change in [O2] is -0.0682 mol/L
- The change in [N2O4] is -0.0341 mol/L
You correctly calculated the change in [N2] and [O2], but your calculation for the change in [N2O4] was incorrect. It should be -0.0341 mol/L, not 0.1021 mol/L.
I hope this clarifies your doubts. Let me know if you have any further questions!