Kc for the equilibrium below is 0.0036 at 2680K. If 1 mole of N2 and 1 mole of 02 are allowed to reach equilibrium at 2680K, what mass of NO will be present in the mixture?
(you will need to solve a quadratic equation to answer this)
N2 + O2 <--> 2NO (reversible reaction)
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- Thread Starter
- 17-01-2017 12:15
- 17-01-2017 15:49
Let this reaction take place in a container of 1dm^3.
Let x be the number of moles of NO in the container at equilibrium at 2680K
For every mole of NO there is, there must be 0.5moles less N2 and 0.5 moles less O2, so since it starts off with 1 mole of O2 and 1 mole of N2, at equilibrium there will be (1-0.5x) moles of N2 and (1-0.5x) moles of O2.
Therefore you the equilibrium constant can be written:
0.0036 = x^2 / (1-0.5x)(1-0.5x)
0.0036 = x^2 / (0.25x^2 - x + 1)
0.0009x^2 - 0.0036x + 0.0036 = x^2
0.9991x^2 + 0.0036x - 0.0036 = 0
Use quadratic equation
x = 6/103
6/103 moles of NO = 6/103 * 28 = 1.63g (3sf)
I believe that's right, but someone might have to correct me.
- 17-01-2017 18:05
Worth point out that the response is correct except the mass of NO is 30 not 28
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- 17-01-2017 19:20