Use data from Appendix IIB in the textbook to calculate the equilibrium constants at 25∘C for each reaction. 2NO(g)+O2(g)⇌2NO2(g)

asked by @kathrynp1 • over 1 year ago • Chemistry • 5 pts

I think we are supposed to use the equation deltaG=-RTlnK

I calculated deltaG= products-reactants but I'm not coming up with the correct answer. Help please

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5 answers

Hey Kathryn! Do you mind sending me a picture of the Appendix IIB? It all depends on what's given there. If not, just send me the values given in the book and the question. Thanks!!

answered by @sabrina • over 1 year ago

deltaGf values:

NO= 86.58 O2: 0

NO2: 51.23

answered by @kathrynp1 • over 1 year ago

Okay, so you are correct, you will be using that equation to solve for K. Remember K is you equilibrium constant, so I rearranged the equation for you in order to solve for K. That's how it should look. Again, you calculate ΔG° by using the formula :

ΔG°rxn = ΔGproducts - ΔGreactants

For your R, you will use the constant R=8.314 J/ mol . K

For your temperature, you must use Kelvin, so add 273 to the 25° C that are given in the problem. You should get T = 298° K.

Just plug all of these values into your calculator and you should get your K. Let me know if this helps!

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answered by @sabrina • over 1 year ago
  • so i have the same question just that the temp is at 500 k but i got 5.49x10^53 and the answer is 1.24x10^50, idk what im doing wrong Celica commented 6 months ago

I'm getting 2.47e12, but it is saying it's the wrong answer.

Here is my work:

2NO: 286.58=173.16 O2:0 2NO2: 251.23=102.46

102.46-173.16=-70.70 converted to J= -70700 J

plug into equation: lnk=-70700/8.314*298)=-28.54

raised to e^= 2.47e12

answered by @kathrynp1 • over 1 year ago

Hey! So the only thing I did different was using 298.15 as my temperature, which didn't make much of a difference, but I got 2.436 x 10 ^ 12. The problem is set up correctly, there might just be an error in the numbers for DeltaG.

answered by @sabrina • over 1 year ago