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?
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 = 4X^{2}
⟹ 4.X^{2} + 6.66×10^{−6}. X - (3.26 x 10^{-5} )= 0
The above equation is in the form of ax^{2} + bx + c = 0
where, X = [(-b) +/- (b^{2} - 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
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