You can work this out with numbers or with reasoning. Numbers take longer but may lead to some more confidence.
Kc = (B)^5/(A)^6
So if pressure decreases you know the reaction will shift to the left. That means Qc will be too large meaning B is too large and A is too small. Therefore, Qc will be 2.40*Kc.
To do it with numbers.
Kc = (0.1042)^5/(0.1887)^6 = 0.272 if I didn't goof with the calculator. YOu should confirm these numbers.
If P goes down by factor 2.40 then V goes up by 2.40 and concn goes down by 2.40. So new A = 0.1887/2.40 = about 0.08 (but that's an estimate and you should get a better number than that) while B changes by 0.1042/2.40 = about 0.04--again an estimate).
Qc = (new B)^5/(new A)^6 = (0.04/0.08) = 0.653 (the end number I cam up with).
You can see that 0.653 is 2.40*0.272 and that is Qc = 2.40*Kc
Check my thinking.
Method for getting correct answer? Thank you!
Initially there were equilibrium concentrations of 0.1887 M A, 0.1042 M B. Which of the equations below is correct if the pressure decreased by a factor of 2.40? Use the reaction below.
6A(g) ↔ 5 B(g)
Qc = 0.42 Kc
Qc = 0.45 Kc
Qc = 2.40 Kc
Qc = 0.38 Kc
2 answers
Thank you! .434^5 / .0786^6 equals .653, that is correct! I was just able to answer another problem similar to this thanks to your help. Many thanks!