Question

Question #9caae

Answer 1

`K_c = 5.81 xx 10^(-3)`

Explanation:

We're asked to find the equilibrium constant `K_c` for this reaction, given an initial and final equilibrium concentration.

The equilibrium constant expression for this reaction is

`K_c = (["SO"_2]^2["O"_2])/(["SO"_3]^2)`

If `0.840` moles of `"SO"_3` is placed in a `4.50`-`"L"` container, the initial concentration is

`["SO"_3] = (0.840color(white)(l)"mol")/(4.50color(white)(l)"L") = 0.187M`

We can tabulate now our initial concentrations for each species:

Initial:

• `"SO"_3`: `0.187M`

• `"SO"_2`: `0`

• `"O"_2`: `0`

The changes in concentration, which we'll call the variable `x`, can be expected via the coefficients of the equation:

Change:

• `"SO"_3`: `-2x` (coefficient of 2)

• `"SO"_2`: `+2x` (coefficient of 2)

• `"O"_2`: `+x` (coefficient of 1)

We know the final equilibrium concentration of `"O"_2` is

`["O"_2] = (0.130color(white)(l)"mol")/(4.50color(white)(l)"L") = 0.0289M`

Therefore, the variable `x` must be equal to this value, because the oxygen concentration change is "`+x`", and hence `0 + x = 0.0289`

Using this value, we can now find the equilibrium concentrations of all species:

Equilibrium:

• `"SO"_3`: `0.187M - 2(0.0289M) = 0.129M`

• `"SO"_2`: `0 + 2(0.0289M) = 0.0578M`

• `"O"_2`: `0 + 0.0289M = 0.0289M`

And lastly, we can plug in these values to the equilibrium constant expression to calculate `K_c`:

`K_c = ((0.0578M)^2(0.0289M))/((0.129M)^2) = color(blue)(5.81 xx 10^(-3)`