1. 2.00 x 10 -4 mol of NH 3 , 1.50 x 10 -5 mol N 2 , and 3.54 x 10 -1 mol H 2 ar

Question

1. 2.00 x 10-4 mol of NH3, 1.50 x 10-5 mol N2, and 3.54 x 10-1 mol H2 are mixed in a 1.00 L rigid container at 500oC. At this temperature, Kc for this reaction is 6.0 x 10-2.

Since Q is less than K, the reaction will shift to the right.

Since Q is greater than K, the reaction will shift to the left.

Since Q equals K, the system is at equilibrium, and no shift will occur.

4. Which of the following will cause a shift in equilibrium to the right for this reaction?

remove PCl3

remove PCl5

decrease the volume of the container

none of the above

5. For the following endothermic reaction at equilibrium, choose the change that will increase the value of K.

increase temperature

remove O2

add SO2

decrease temperature

Since Q is less than K, the reaction will shift to the right.

Since Q is greater than K, the reaction will shift to the left.

Since Q equals K, the system is at equilibrium, and no shift will occur.

Explanation / Answer

There are multiple questions here. . i am allowed to answer only 1 at a time. I will answer question 1 for you.Please ask other as different question.

1.
N2 + 3H2 <---> 2NH3

Qc = [NH3]^2 / {[N2] [H2]^3}
        = (2*10^-4)^2 / {(1.5*10^-5) * (3.54*10^-1)}
       = 7.53*10^-3

But Kc is 6*10^-2

Since QC< Kc, more product needs to be formed and gence reaction will shift right

Answer: a

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