At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M and [NO]=0.500 M....

At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M and [NO]=0.500 M.

N2(g)+O2(g)<----->2NO(g)

If more NO is added, bringing its concentration to 0.800 M, what will the final concentration of NO be after equilibrium is re-established?

____________M

Solutions

Expert Solution

the first thing to do is to calculate the equilibrium constant of concentration Kc so:

N2(g)+O2(g)<----->2NO(g).

the equilibrium constant is the relationship between products and reactants so:

Kc = [NO]² / [N₂] [O₂]

calculate Kc with the given numbers:

Kc = [0.5]² / [0.2] [0.2] = 6.25

now analyze the equilibrium with the concentration of NO = 0.8 M, according to lecheteliers principle the equilibrium will shift to the left because you have more of product than the original case

N2(g) + O2(g)<-----> 2NO(g).

I........0.2...............................0.2..........................0.8

C.....+x................................+x............................-x

E......0.2+x...........................0.2+x....................0.8-2x

the equilibrium constant is expressed as:

(0.8 - 2x)² / (0.2+x)(0.2+x) = 6.25

solve this equation to get the value of x, i am using a spreadsheet so x is 0.0666 M

the concentration of NO is

0.8 - 2 * 0.0666 = 0.666 M

*hope it helps =)

queen_honey_blossom answered 1 year ago

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