Question

A(aq) <--> 3B(aq)

Kc = 3.8 x 10^-6 at 500K

If a 2.40 M sample of A is heated to 500 K, what is the concentration of B at equilibrium?

Please write out the steps so I can understand it.

Answer 1

Ans--- Given reaction is , A(aq) <===> 3B(aq)

given , initial concentration of A is 2.40 M and kc=3.8* 10^-6 , we have to calculate concentration of B at equilibrium . for calculating concentration of B , we can use the ICE table as follows:

  A(aq) <===> 3B(aq)

initial concentration (I) ............... 2.40 0

change of concentration (C) .................. -X + X

Equilibrium concentration (E) .................. (2.40-x) 3x

from the above reaction we can write the Kc as

Kc=[B]³ / [A] ---------(1)

3.8 * 10^-6 = (3X) ³ / (2.40-X)

   3.8 * 10^-6 = 27X ³ / (2.40-X) [As the value of kc is very small , so x in the denominator is neglected]

   27X ^{3 ₌} 3.8 * 10^-6(2.40)

   27X ^{3 ₌} 9.12* 10^-6

x³=0.337 *10^-6

   x³=3.37 *10^-7

X= 0.0069

hence concentration of B at equilibrium is [B]=0.0069M