Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a tank with of ammonia gas, and when the mixture has come to equilibrium measures the amount of nitrogen gas to be . Calculate the concentration equilibrium constant for the decomposition of ammonia at the final temperature of the mixture. Round your answer to significant digits.

Kc = 15.25

Explanation:

In this case, As I say in the comments, you are missing the following data of moles and volume which are:

Initial moles of ammonia = 29 mol

Final moles of nitrogen = 13 mol

Volume tank = 75 L

Now, if your data is different, just replace the values and follow the same procedure written here to get the accurate result.

With these data, let's write the equilibrium reaction that is taking place here:

2NH₃ <> N₂ + 3H₂     Kc = ?

To get the Kc, we need to use the following expression:

Kc = [N₂] [H₂]³ / [NH₃]²

This expression is only usable when we reach the equilibrium. But we don't have the concentrations in equilibrium of any species, so, we need to get those values. To get that, we need to do an ICE chart, write the initial and final concentrations given, and then, replace the values.

First, the concentrations of the species are:

[NH₃] = 29 / 75 = 0.3867 M

[N₂] = 13 / 75 = 0.1733 M

Now that we have these values, we will replace them in the ICE chart, and then, rewrite the Kc expression to get that value so:

       2NH₃ <> N₂ + 3H₂     Kc = ?

i)     0.3867               0        0

c)       -2x                 +x      +3x

e)   0.3867-2x          x         3x

However, as we already have the value for concentration of nitrogen, this will be the value of "x", therefore, our final values in equilibrium would be:

[NH₃] = 0.3867 - 2(0.1733) = 0.0401 M

[H₂] = 3(0.1733) = 0.5199 M

Now with these values, we can replace them in the Kc expression to get that value so:

Kc = (0.5199)³ * (0.1733) / (0.0401)²

Kc = 0.0244 / 0.0016

Kc = 15.25

And this is the value of Kc for this reaction.

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )