A researcher is studying the decomposition of A as shown by the general reaction below:
2 A(g) 4 B(g) + 2 C(g)
Initially, the scientist fills an evacuated 1.772 L flask with 3.710 x 10-1 moles of species A. Upon equilibrium, it is determined that the concentration of A is 4.606 x 10-2 M. Calculate Kc.
This is what I have done:
2A ---> 4B + 2C
I: 3.710E-2 mol --> 0 M + 0 M
2.094E-1 M ---> 0 M + 0 M
C: -1.633E-1 ---> +_______ + ________
E: 4.606E-2 M ---> _______ + ________
Kc= ([B]^4[C]^2)/((4.606E-2)^2)
You're ok with 0.2094M to start and 1.633 M change and 0.04606 Equil but it stops there. Here is what you do. By the way, note how I get around the spacing problem.You also need to note the arrow.
............2A ==> 4B + 2C
initial..0.20937....0.....0
change......-2x.....4x....2x
equil......0.04606..........
2x = 0.20937-0.04606 = 0.16331
Therefore, x = 0.16331/2 = 0.08165M
Then equil B = 4x and equil C = 2x.
Substitute those into Kc expression and solve.
I carried more places than allowed by the significant figure rules; you can round as your instructor does. I round at the end.
Thank you Dr. Bob.
To calculate Kc, the equilibrium constant, you need to determine the concentrations of species B and C at equilibrium.
From the balanced equation, you can see that for every 2 moles of A that decompose, you get 4 moles of B and 2 moles of C. So the change in concentration for B and C will be directly related to the change in concentration of A.
Given that the initial concentration of A is 3.710 x 10^-1 moles and the equilibrium concentration of A is 4.606 x 10^-2 M, you can calculate the change in concentration of A:
Change in A = Initial concentration of A - Equilibrium concentration of A
= (3.710 x 10^-1) - (4.606 x 10^-2)
= 2.249 x 10^-1
Since the change in A is equal to the stoichiometric coefficient for A in the balanced equation (2 moles), you know that the change in concentration of B and C will be twice the change in concentration of A.
Change in B = 2*(Change in A) = 2*(2.249 x 10^-1) = 4.498 x 10^-1
Change in C = 2*(Change in A) = 2*(2.249 x 10^-1) = 4.498 x 10^-1
Now you have the equilibrium concentrations of all the species:
[B] = Change in B / Total Volume = (4.498 x 10^-1 / 1.772 L) = 2.538 x 10^-1 M
[C] = Change in C / Total Volume = (4.498 x 10^-1 / 1.772 L) = 2.538 x 10^-1 M
Finally, you can calculate Kc using the equation you mentioned:
Kc = [B]^4[C]^2 / [A]^2
= (2.538 x 10^-1)^4 * (2.538 x 10^-1)^2 / (4.606 x 10^-2)^2
= 2.576 x 10^3
Therefore, the value of Kc for this reaction is 2.576 x 10^3.