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
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.
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!
To determine which equation is correct, we need to calculate the new equilibrium concentrations of A and B after the pressure has decreased by a factor of 2.40.
To do this, we can use the relationship between the initial concentrations, the equilibrium concentrations, and the reaction's stoichiometry.
Here's how we can calculate the new equilibrium concentrations:
1. Start by writing down the balanced chemical equation:
6A(g) ↔ 5B(g)
2. Use the given initial equilibrium concentrations:
[A]₀ = 0.1887 M
[B]₀ = 0.1042 M
3. Calculate the reaction quotient (Qc) using the initial concentrations:
Qc = [B]₀^5 / [A]₀^6
4. Since the pressure has decreased by a factor of 2.40, we need to adjust the initial concentrations accordingly:
[A] = [A]₀ / 2.40
[B] = [B]₀ / 2.40
5. Calculate the new reaction quotient (Qc) using the adjusted concentrations:
Qc = [B]^5 / [A]^6
6. Compare the calculated Qc with the given Kc values to determine the correct equation. The equation with the Qc value that matches one of the given Kc values is the correct equation.
Now, let's calculate the new equilibrium concentrations and the Qc value for each given equation:
For equation Qc = 0.42 Kc:
Substitute the adjusted concentrations into the Qc expression:
Qc = ([B] / [A])^5
= ([0.1042 M / 2.40] / [0.1887 M / 2.40])^5
≈ 0.2870
For equation Qc = 0.45 Kc:
Substitute the adjusted concentrations into the Qc expression:
Qc = ([B] / [A])^5
= ([0.1042 M / 2.40] / [0.1887 M / 2.40])^5
≈ 0.3021
For equation Qc = 2.40 Kc:
Substitute the adjusted concentrations into the Qc expression:
Qc = ([B] / [A])^5
= ([0.1042 M / 2.40] / [0.1887 M / 2.40])^5
≈ 0.1274
For equation Qc = 0.38 Kc:
Substitute the adjusted concentrations into the Qc expression:
Qc = ([B] / [A])^5
= ([0.1042 M / 2.40] / [0.1887 M / 2.40])^5
≈ 0.2627
Comparing the calculated Qc values with the given Kc values, we find that none of the calculated Qc values match any of the given Kc values. Therefore, none of the equations (Qc = 0.42 Kc, Qc = 0.45 Kc, Qc = 2.40 Kc, or Qc = 0.38 Kc) are correct.
Hence, based on the calculations, none of the equations are correct.