Respuesta :
Answer:
The concentrations of A, B, and C at equilibrium:
[A] = 0.0 M
[B] = 2.7 M
[C] = 2.4 M
Explanation:
Concentration of 1.80 mol of A in 1.00 L container :
[tex][A]=\frac{1.80 mol}{1.00 L}=1.80 M[/tex]
Concentration of 3.90 mol of B in 1.00 L container :
[tex][B]=\frac{3.90 mol}{1.00 L}=3.90 M[/tex]
[tex]3A(g)+2B(g)\rightleftharpoons 4C(g), K_c=2.13\times 10^{17}[/tex]
Initially
1.80 M 3.90 M 0
At equilibrium
(1.80-3x)M (3.90-2x) 4x
The expression of an equilibrium constant is given by :
[tex]K_c=\frac{[C]^4}{[A]^3[B]^2}[/tex]
[tex]2.13\times 10^{17}=\frac{(4x)^4}{(1.80-3x)^3\times (3.90-2x)^2}[/tex]
Solving for x:
x = 0.600
The concentrations of A, B, and C at equilibrium:
[A] = [1.80-3x]=[1.80-3 × 0.600]= 0 M
[B] = [3.90-2x] = [3.90-2 × 0.600] = 2.7 M
[C] = [4x] =[4 × 0.600 M] = 2.4 M
