In a one litre flask, 5.6 x 10^-6 mol of A, & 5 x 10^-5 mol of B are mixed. At equilibrium, there is 4.8 x 10^-5 mol of B. Calculate the value of the equilibrium constant.

Question

daenio · Answer

The equation of the system is 2A$$2A _{(g)} + B _{(g)}\rightarrow 3C _{g)}$$

.sam. · Answer

At equilibrium, there is 4.8 x 10^-5 mol of B or C?

daenio · Answer

B.

daenio · Answer

@.Sam.

.sam. · Answer

Just to verify my answer, do you have the answer?

.sam. · Answer

I'll give it a try, the equation that they gave you is 
$$2A _{(g)} + B _{(g)}\rightarrow 3C _{g)}$$
We can construct an ICE table to get things clear. So,
The volume is 1Liter=1dm3, so the concentration of solutions will be the same.

                     2A             +           B     ------->        3C
I            5.6x10^-6                 5x10^-5                    0
C              -2x                              -x                        +3x 
E            1.6x10^-6                 4.8x10-5               6x10^-6
We have to lose some reactants to gain products since there's no product to start with.
Since at equilibrium, there is 4.8 x 10^-5 mol of B, we use this to find the final concentration of both A and C.

The equilibrium expression is given by
$$K_c=\frac{[C]^3}{[A]^2[B]}$$
Substitute in and solve for $K_c$