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.
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)
2 answers
Thank you Dr. Bob.