Question

Consider the following reaction. A(aq) <---> 2B(aq) Kc=8.36 x 10^-6 at 500K If a 3.80 M...

Consider the following reaction. A(aq) <---> 2B(aq) Kc=8.36 x 10^-6 at 500K

If a 3.80 M sample of A is heated to 500 K, what is the concentration of B at equilibrium?

Homework Answers

Answer #1

A(aq) <---> 2B(aq)

initial concentration 3.80 M 0

concentration at 3.80 - 2x 2x

equillibrium

hence K (equillibrium constant is iven by )

K = [B]2 / [A]      = [2x]2/ [3.80 - 2x ]

8.36 x 10-6 (3.80 - 2x) - 4x2=0 or   4x2 + 16.72x*10-6 - 31.76 * 10-6 = 0

using quadratic equation =

= -b +- (b2 - 4ac ) /2a = - 16.72 +-(279.55*10-12 - 4 *16.72 *31.76 * 10-6 )/ 8

=  - 16.72*10-6 +- (279.55*10-12- 2124.10 *10-6) / 8 (279.55*10-12is small and neglected in comparison

= - 16.72*10-6 +- 2124.10 *10-6 /8 to 2124.10 *10-6)

= - 16.72*10-6 + 46 .08 * 10-3 /8 ( -ve value of square root is neglected )

= 46.06 * 10-3/8 = 5.75 * 10-3 = x

therefore the concerntration of B = 2x= 11.5 * 10-3 M

  

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