Question

Thank you both!

Answer 1

A) is correct
C) is \(K_p' = 1/\sqrt {K_p}\)

Answer 2

Thanks!
Alright so for "B" was this incorrect?
I plugged my original answer back into the equation, but I thought that's how to check. That's where I became confused.

Answer 3

Sorry, as I have not done chemistry for many years, I am not sure anymore how to solve question B).

Answer 4

For B) use the relation: \(K_p=K_c(RT)^{\Delta n}\)

\(K_c=\dfrac{K_p}{(RT)^{\Delta n}}=\dfrac{1.45*10^{-5}}{[(0.0821L*atm/mol*K)(500+237)K]^{(2-4)}}=0.05308\)

Answer 5

I wrote 237 instead of 273, accidentally.

the answer should be 0.0584

Answer 6

Thanks bro!

Alright, so I'd like to double check for Part C I came up w/

\[k_p = \frac{ 1 }{ (1.45*10^-5)^\frac{ 1 }{ 2 } } = 262.6 \approx 263\]
I solved slightly different than Vincent's formula, but mathematically it's the same.
Did I solve correctly?

Answer 7

yes, it's the exact same thing he wrote, because \(\Large \sf \sqrt x =x^{\frac{1}{2}}\)

Answer 8

Thanks my friends!