Consider the reaction. a(aq) = 2 b(aq) kc = 6.90 x 10 6 at 500 k if a 3.00 m sample of a is heated to 500 k, what is the concentration of b at equilibrium?
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Ответ:
0.004548 M is the concentration of B at equilibrium at 500 K.
Explanation:
A(aq) ⇆ 2 B(aq)
Initially 3.00 M
At equilibrium 3.00 -x 2x
Equilibrium constant of the reaction at 500 K =![K_c=6.90\times 10^{-6}](/tpl/images/0224/5358/30840.png)
Concentration of A at 500 K at equilibrium , [A] = (3.00 -x )M
Concentration of B at 500 K at equilibrium,[B]= 2x
An expression of equilibrium constant is given as:
On solving for x:
x = 0.002274 M
[B] = 2 x = 2 × 0.002274 M = 0.004548 M
[A] = (3-x) = 3 M - 0.002274 M =2.997726 M
0.004548 M is the concentration of B at equilibrium.
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