# Question 9caae

Jun 26, 2017

${K}_{c} = 5.81 \times {10}^{- 3}$

#### 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

${K}_{c} = \left({\left[{\text{SO"_2]^2["O"_2])/(["SO}}_{3}\right]}^{2}\right)$

If $0.840$ moles of ${\text{SO}}_{3}$ is placed in a $4.50$-$\text{L}$ container, the initial concentration is

$\left[\text{SO"_3] = (0.840color(white)(l)"mol")/(4.50color(white)(l)"L}\right) = 0.187 M$

We can tabulate now our initial concentrations for each species:

Initial:

• ${\text{SO}}_{3}$: $0.187 M$

• ${\text{SO}}_{2}$: $0$

• ${\text{O}}_{2}$: $0$

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

Change:

• ${\text{SO}}_{3}$: $- 2 x$ (coefficient of 2)

• ${\text{SO}}_{2}$: $+ 2 x$ (coefficient of 2)

• ${\text{O}}_{2}$: $+ x$ (coefficient of 1)

We know the final equilibrium concentration of ${\text{O}}_{2}$ is

$\left[\text{O"_2] = (0.130color(white)(l)"mol")/(4.50color(white)(l)"L}\right) = 0.0289 M$

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:

• ${\text{SO}}_{3}$: $0.187 M - 2 \left(0.0289 M\right) = 0.129 M$

• ${\text{SO}}_{2}$: $0 + 2 \left(0.0289 M\right) = 0.0578 M$

• ${\text{O}}_{2}$: $0 + 0.0289 M = 0.0289 M$

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)#