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November 26, 2014

November 26, 2014

Posted by **Summer** on Friday, December 9, 2011 at 8:08pm.

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)

- Chemistry -
**DrBob222**, Friday, December 9, 2011 at 10:19pmYou'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.

- Chemistry -
**Summer**, Friday, December 9, 2011 at 10:43pmThank you Dr. Bob.

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