# Equilibrium question Destroyer #67

Discussion in 'DAT Discussions' started by Troyvdg, Jul 27, 2011.

1. ### Troyvdg Dentistry not Debtistry 7+ Year Member

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consider the following EQ.

4A (s) + 2 B (g) -- 4 C (g)

K=50

which is false:
a) K= [C]^4/{B]^2

b) K = &#8730;1/50 for 2C (g) -- 2B (g) + 2A (s)

c) changing the answer will not change K

d) 2(50) for 8A (s) + 4 B (g) -- 8C (g)

I understand that A is true, but for B (which is also true) I understand why it is the inverse of Keq, because its the reverse reaction, but why do we take the square root?

and for D, the correct answer would have been (50)^2 but why?

I'm not understanding destroyer's explanation. Thanks!

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Last edited: Jul 27, 2011
2. ### mh0000

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Whenver you multiply a balanced equation by some factor (all of the terms must be multiplied by this factor, not just some), just raise the equilibrium constant (K) by that factor.

A + B --> C ; K=[C]/[A]

Multiply eqn by a factor of 2, raise the K by 2:
2(A + B --> C) = 2A + 2B --> 2C ; K=([C]/[A])^2

Likewise, halving the eqn coefficients is the same as multiplying by 1/2
(1/2)(A + B --> C) = 0.5A +0.5B --> 0.5C ; K=([C]/[A])^(1/2) or sqrt([C]/[A])

3. OP

### Troyvdg Dentistry not Debtistry 7+ Year Member

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awesome. thank you

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