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?
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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