Question

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?

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