5J.13 and Ideal Gas Law

[Alice Chang 2H]

I'm having a bit of trouble trying to figure problem 5J.13 out. I'm trying to work backwards with the given answer for this question, but I'm still getting stuck.

In the problem, the concentration can be worked out, and the temperature is given, so the partial pressures (?) of each molecule can be worked out (or at least I'm assuming by the given units of bar). Yet when I use P=(n/V)RT, I still can't get the numbers the answer got for the conversion from concentration to bars. What am I doing wrong? Am I using the wrong formula?

[Daniel Honeychurch1C]

You don't need to do any calculations, just look at the equilibrium constants for 600 K and 700 K. The Keq is smaller at the higher temperature, so increasing the temperature will increase the concentration of the reactants. Therefore, the concentration of NH3 will decrease.

[cassidysong 1K]

There is no need to calculate anything; just compare the two different K values.

[Asia Yamada 2B]

Yes, you don’t have to calculate anything in this problem. All you have to do is look at the two K values and use Le Chatelier’s Principle. When the equilibrium mixture is heated, the K value decreases, so that means there’s more reactants than products and there will be less ammonia present since it’s a product.

[Natalie Do 3F]

Raising temperature favors reactants
This means there will be less ammonia at higher temperatures

All of this is assuming no other changes are occurring

[Bai Rong Lin 2K]

You would just look at the difference of the two K value.

[Kiyoka Kim 3C]

The data provided by the question tells you that raising the temperature leads to a smaller K value which means at equilibrium [R]>[P] relative to the lower temperature. Thus the reverse reaction is favored and less ammonia will be present at higher temperatures.

[305572629]

Just compare the two K values given. I was confused, too, until I realized I didn’t need to calculate any other values.

[Lorena_Morales_1K]

what exactly is the ideal gas law?