Chemical Equilibrium (answer check)

[ScrubsFan]

Consider the following reaction

A(g) <---> 2B(g) + C(g)

When 1.00 mol of A is placed in a 4.00L container at temperature t, the concentration of C at equilibrium is 0.050 mol/L. What is the equilibrium constant for the reaction at temperature t?

Here is my answer:

The balanced equation shows that for each mol of C formed, 2 mol of B is formed and 1 mol of A is consumed. We are told that, at equilibrium, 0.050 mol of C has been produced. This means that 0.050 mol of B must also have been produced and that 0.050 mol of A was consumed. Therefore, at equilibrium,

0.050 mol C x 2mol B / 1mol C = 0.1 mol B

1.00 mol A - [0.050 mol C x 1mol A / 2mol C]

= 0.975 / 0.98 mol A

and the equilibrium concentrations are:

[C] = 0.050 mol / 4.00 L = 0.0125 mol/L

2 = 0.1 mol / 4.00 L = 0.025 mol/L

[A] = 0.98 mol / 4.00 L = 0.245 mol/L

Substitute the equilibrium concentrations in the equilibrium law expression and solve for Ke:

Ke = 2[C] / [A]

= 0.025 mol/L x 0.0125 mol/L
0.245 mol/L

= 0.0013 mol/L

 

[Bystander]

0.050 mol C x 2mol B / 1mol C = 0.1 mol B

It goes "south" from here; mixed stoichiometric and concentrative expressions. Salvageable? Maybe.

 

[epenguin]

You're given final [C]. → [ B]

[C] → moles C. Moles C, initial moles A, → equilibrium moles A. Is in 4 L → [A]

Then you have all you need to calculate K.

I made it 0.4.

 

[Bystander]

[A]_Init = 0.25: [A]_Final = 0.2; (B)_Final = 0.1, and [C]_Final = 0.05, and 5 x 10^-4 over 2 x 10^-1 = 2.5 x 10^-3?

OK, how do I get conc. of B to render?

 

[epenguin]

BTW OP's expression for the equilibrium law and constant is wrong.

 

[Bystander]

A(g) <---> 2B(g) + C(g)

BTW OP's expression for the equilibrium law and constant is wrong.

The problem statement itself?