# chemistry

Calculate the equilibrium concentrations of all three substances.

Kc=1.7e-3
Q=3.6e-3

[NO]=0.0015 mol/L
[O2]=0.025 mol/L
[N2]=0.025 mol/L

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1. With Q>K it means products are too large and reactants too small which means the reaction will move to the product side to achieve equilibrium.

You need to show the reaction to know which Kc we are talking about; i.e., the forward or reverse reaction.

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2. the formula is:
N2(g)+ O2(g) -><- 2NO(g)

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posted by Lauren
3. I assume the concentrations shown are initial concentrations.
...........N2(g)+ O2(g) <--> 2NO(g)
I.........0.025..0.025.......0.0015
C..........+x......+x........-2x
E.......0.0025+x..0.025+x....0.0015-2x

Kc = (NO)^2/(N2)(O2)
Substitute into Kc expression and solve for x, the evaluate 0.025+x and 0.0015-x.
Post your work if you get stuck.
Note: If you are careful you may not be required to solve a long quadratic equation.

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