When N2(g) (2.983 mol) and 95.46 grams of O2(g) in a 150.0 L reaction vessel at 782.0 K are allowed to come to equilibrium...
the mixture contains 99.37 grams of NO(g). What is the concentration (mol/L) of O2(g)?
N2(g)+O2(g) = 2NO(g)
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Since this is an equilibrium problem, we need to set up an ICE table. Since we need concnetrations of each species, we'll have to do some conversions of the amounts to find molarity. The temperature isn't really part of the problem other than knowning it's constant. For N2, it's just dividing 2.983 mol/150.0 L to get 0.01989 M.
For oxygen, we need to convert to moles, then to molarity
95.46 g (1 mol / 32.0 g) = 2.983 mol
0.2983 mol O2 / 150.0 L = 0.01989 M O2
These two values serve as our initial concentration and NO starts at 0. Then we can put the change in as a function of x based on the stoichiometry of the balanced equation and find the equilibrium concentrations.
N2 O2 2NO
I 0.01989 0.01989 0
C -x -x +2x
E 0.01989-x 0.01989-x 2x
Now, the problem tells us that we end up with 99.37 g of NO which we can convert to moles, then to molarity which will be the equilibrium concnetration of NO and therefore will be equal to 2x.
99.37 g (1 mol / 30.0 g) = 3.312 mol
3.312 mol / 150.0 L = 0.02208 M
So, we now know that
2x = 0.02208 M
x = 0.01104 M
Since we know the value of x we can find the final concentration of O2 by subtracting
0.01989 - 0.01104 = 0.00885 M O2
which will also be the same concentration as N2 since they had the same initial concentration and both lost "x".
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