Question

Consider the reaction. A(aq)↽−−⇀2B(aq)?c=6.66×10−6 at 500 K If a 4.90 M sample of A is heated to 500 K, what is the concentration of B at equilibrium?

Answer 1

Considering the reaction. A(aq) ⇌ 2B(aq) ?_c = 6.66×10⁻⁶ we use the ICE table to get concentration of B at equilibrium

	A(aq)	⇌	2B(aq)
Initial	4.90 M		0
Change	- X		+ 2X
Equilibrium	(4.90 - X )		2X

The Equilibrium Constant,

?_c = [B(aq) ]² / [A(aq)]

⟹ 6.66×10⁻⁶ =  (2X )²/ (4.90 - X )

⟹ (6.66×10⁻⁶ ) (4.90 - X ) = 4X ²

⟹ 3.26 x 10^-5 - 6.66×10⁻⁶ x X = 4X²

⟹ 4.X² + 6.66×10⁻⁶. X - (3.26 x 10^-5 )= 0

The above equation is in the form of ax² + bx + c = 0

where, X = [(-b) +/- (b² - 4.a.c)^(1/2)] / 2.a

for a = 4 , b = 6.66×10⁻⁶ , c = - 3.26 x 10^-5 on solving for X we get

⟹ X = 0.0029 or  - 0.0029

At taking the positive value of X at equilibrium [B] = 2 X M = 2 x 0.0029 M = 0.0058 M  

Hence , the concentration of B at equilibrium is [B] = 0.0058 M