Answer:
[tex][C]_{eq}=4(0.733M)=2.932M[/tex]
[tex][A]_{eq}=2.20M-3(0.733M)=0.001M[/tex]
[tex][B]_{eq}=3.70M-2(0.733M)=2.23M[/tex]
Explanation:
Hello,
In this case, probably, the undergoing chemical reaction is:
[tex]3A(g)+2B(g)<-->4C(g)[/tex]
Nonetheless, the equilibrium constant is not given, but could be assumed, therefore, let me assume it is 1.13x10¹⁹ (very close to a constant I found for a similar exercise on the Ethernet), thus, the initial concentrations of A and B are computed below based on the given moles and the container volume:
[tex][A]_0=\frac{2.20mol}{1.00L} =2.20M[/tex]
[tex][B]_0=\frac{3.70mol}{1.00L} =3.70M[/tex]
In such a way, by applying the law of mass action in addition to the ICE table, the equilibrium statement turn out into:
[tex]Kc=\frac{[C]_{eq}^4}{[A]_{eq}^3[B]_{eq}^2}[/tex]
Now, by inserting the initial concentrations of both A and B and the reaction change, [tex]x[/tex], one obtains:
[tex]1.13x10^{19}=\frac{(4x)^4}{(2.20M-3x)^3(3.70M-2x)^2}[/tex]
Such equation is a highly non-lineal one, that is why we must apply a method such as Newton-Raphson to compute the roots which define the value of the change [tex]x[/tex], thus, after applying it, the roots are:
[tex]x_1=0.733\\x_2=1.85[/tex]
The valid root is [tex]x_1=0.733[/tex] as the other one produces a negative concentration of A at equilibrium, therefore, the requested concentration turn out into:
[tex][C]_{eq}=4(0.733M)=2.932M[/tex]
[tex][A]_{eq}=2.20M-3(0.733M)=0.001M[/tex]
[tex][B]_{eq}=3.70M-2(0.733M)=2.23M[/tex]
However, you can just modify Kc if is different to the one I assumed and perform the Newton-Raphson method to compute the adequate root.
Best regards.
