# Chemistry

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SO this is my first time doing this... lol i need help on an AP chemistry question for equilibrium.

A 0.500 L tank contains 3.00 g of NO(g) at 750. K. The equilibrium constant for the reaction
below at this temperature is 3.4 x 10
-3
2NO(g) ⇌ N2(g) + O2(g)

(a) write the equilibrium constant expression for this reaction:

(b) calculate Kc for the reaction 2N2(g) + 2O2(g) ⇌ 4NO(g)

(c) determine the equilibrium concentrations of all species

• Chemistry -

moles NO = 3.00g/molar mass = 3.00/30 = 0.1 mol.
a. I assume you can do this. It's just products/reactants with coefficients becoming exponents.

b. This reaction is doubled and reversed. Take the reciprocal of the original Kc to take care of the reversed part and square everything to take care of the doubled part.

c. I assume this is for the original K from part a.
.........2NO ==> N2 + O2
initial.0.1......0....0
change..-2x......x.....x.
equil...0.1-2x....x.....x

Substitute the equilibrium values from the ICE chart into Kc expression (the one you wrote in part a) and solve for x and 0.1-2x.

• Chemistry -

That's wrong