Suppose a 500.mL flask is filled with 1.6mol of NO3 and 1.2mol of NO2 . The following reaction becomes possible: +NO3gNOg 2NO2g The equilibrium constant K for this reaction is 8.01 at the temperature of the flask. Calculate the equilibrium molarity of NO3 . Round your answer to two decimal places.

At equilibrium, the concentration of is 1.60 M

Explanation:

Initial values

We have the balanced chemical reaction is:

+ ⇄2

:

:

Equilibrium values

and  are going to consume, while is going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: []=3.2M   ;     []= 2.4M    ; []=0

C:  []=-x   ;     []= -x   ; []=+2x

E:  []=3.2-x  ;     []= 2.4-x    ; []=0+2x

Now we can use the constant information:

K=8.01

Solving for x we have 1.6 and 9.6 as possible results. But 9.6 has no sense because the concentration left can not be negative. So x=1.6

It means that at equilibrium, the concentration of is going to be 1.6 M

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)

is there any options? maybe im just not seeing them but i need the options to

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