Question #9caae

1 Answer
Jun 26, 2017

Answer:

#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)#