Question

At a certain temperature, this reaction establishes an equilibrium with the given equilibrium constant, kc. if, at this temperature, 2.20 mol of a and 3.70 mol of b are placed in a 1.00-l container, what are the concentrations of a, b, and c at equilibrium?

Answer 1

6.9 i  it think so ???????????????????????????????????????????????

Answer 2

`[C]_(eq)=4(0.733M)=2.932M`

`[A]_(eq)=2.20M-3(0.733M)=0.001M`

`[B]_(eq)=3.70M-2(0.733M)=2.23M`

Step-by-step explanation:

Hello,

In this case, probably, the undergoing chemical reaction is:

`3A(g)+2B(g)<-->4C(g)`

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:

`[A]_0=(2.20mol)/(1.00L) =2.20M`

`[B]_0=(3.70mol)/(1.00L) =3.70M`

In such a way, by applying the law of mass action in addition to the ICE table, the equilibrium statement turn out into:

`Kc=([C]_(eq)^4)/([A]_(eq)^3[B]_(eq)^2)`

Now, by inserting the initial concentrations of both A and B and the reaction change,
`x`, one obtains:

`1.13x10^(19)=((4x)^4)/((2.20M-3x)^3(3.70M-2x)^2)`

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
`x`, thus, after applying it, the roots are:

`x_1=0.733\\x_2=1.85`

The valid root is
`x_1=0.733` as the other one produces a negative concentration of A at equilibrium, therefore, the requested concentration turn out into:

`[C]_(eq)=4(0.733M)=2.932M`

`[A]_(eq)=2.20M-3(0.733M)=0.001M`

`[B]_(eq)=3.70M-2(0.733M)=2.23M`

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.