Answer:
0.146 M
Explanation:
Equation for the reaction :
2A(aq) ↔ B(aq) + C(aq)
[tex]K_c[/tex] = 65
Molar concentration of A = [tex]\frac{2.50 mol}{1.00 L}[/tex]
= 2.5 M
2A(aq) ↔ B(aq) + C(aq)
Initial 2.50 0 0
Change - 2x + x + x
Equilibrium 2.50 - 2x +x +x
[tex]K_c =\frac {[B][C]}{[A]^2}[/tex]
[tex]65 = \frac{[x][x]}{[2.5-2x]^2}[/tex]
[tex]65 = \frac{[x]^2}{[2.5-2x]^2}[/tex]
[tex]65 = (\frac{[x]}{[2.5-2x]})^2[/tex]
[tex]\sqrt 65 = \sqrt {(\frac{[x]}{[2.5-2x]})^2}[/tex]
[tex]8.062 = \frac{x}{2.5-2x}[/tex]
8.062(2.5 - 2x) = x
20.155 - 16.124x = x
20.155 = 16.124x+x
20.155 = 17.124x
x = [tex]\frac{20.155}{17.124}[/tex]
x = 1.177
[A] = 2.5 - 2x
= 2.5 - 2(1.177)
= 0.146 M
Therefore, the equilibrium concentration of A = 0.146 M
In the given equilibrium, if 2.50 mol of A is added to prepare 1.00 L of solution, the equilibrium concentration of A will be 0.16 M.
In a chemical reaction, chemical equilibrium is the state in which both the reactants and products are present in concentrations which have no further tendency to change with time, so that there is no observable change in the properties of the system.
There are 2.50 moles of A in 1.00 L of solution.
[A] = 2.50 mol/1.00 L = 2.50 M
2 A(aq) ↔ B(aq) + C(aq)
I 2.50 0 0
C -2x +x +x
E 2.50-2x x x
The equilibrium constant (Kc) is:
Kc = 65 = [B] [C]/ [A]²
65 = x² / (2.50-2x)²
x = 1.33 (neglected cause it would cause a negative concentration) and x = 1.17.
The concentration of A at equilibrium is:
[A] = 2.50 - 2 x = 2.50 - 2(1.17) = 0.16 M
In the given equilibrium, if 2.50 mol of A is added to prepare 1.00 L of solution, the equilibrium concentration of A will be 0.16 M.
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