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August 23, 2014

August 23, 2014

Posted by **Sam** on Monday, April 9, 2012 at 8:56pm.

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 -
**DrBob222**, Monday, April 9, 2012 at 10:26pmmoles 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.

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