How would you calculate K for the following equilibrium when [#SO_3#] = 0.0160 mol/L, [#SO_2#] = 0.00560 mol/L. and [#O_2#] = 0.00210 mol/L?

Question

Serenity Allen · Accepted Answer

Given the following equilibrium: [SO3] = 0.0160 mol/L, [SO2] = 0.00560 mol/L, and [O2] = 0.00210 mol/L, how would you compute K for it?

#K = 3.89 × 10^3#

As "the following equilibrium" is not identified by you, I will presume that it is

#"2SO"_2 + "O"_2 ⇌ "2SO"_3#

additionally

#K = 0.0160^2/(("0.005 60")^2 × "0.002 10") = (2.56×10^"-4")/(3.136 × 10^"-5" × 2.10 ×10^"-3") = 3.89 × 10^3#

In the event that the balance is

#"2SO"_3 ⇌ 2"SO"_2 + "O"_2 #

#K = (["SO"_2]^2["O"_2])/ ["SO"_3]^2 = 1/3890 = 2.57 × 10^"-4"#

Mateo Aguilar · Answer

Here's what I got.

Start by writing out the equilibrium reaction given to you #color(red)(2)"SO"_ (2(g)) + "O"_ (2(g)) rightleftharpoons color(blue)(2)"SO"_ (3(g)) # Now, the equilibrium constant for this equilibrium, #K_c#, is calculated by using the equilibrium concentrations of the chemical species that take part in the reaction -- keep in mind that solids and liquids are excluded from the expression of the equilibrium constant! More specifically, the equilibrium constant is equal to the equilibrium concentration of the product raised to the power of its stoichiometric coefficient divided by the product of the equilibrium concentrations of the reactants, also raised to the power of their respective stoichiometric coefficients. In your case, you have sulfur trioxide, #"SO"_3#, as the product and sulfur dioxide, #"SO"_2# and oxygen gas, #"O"_2#, as the reactants. Notice that the equilibrium concentration of the product is higher than the equilibrium concentrations of the two reactants. This tells you that the above equilibrium lies to the right, which implies that #K_c >1#. You will have #K_c = (["SO"_3]^color(blue)(2))/(["SO"_2]^color(red)(2) * ["O"_2])# All you have to do now is use the date given to you by the problem #K_c = ( 0.0160^color(blue)(2) color(darkgrey)(cancel(color(black)(("mol L"^(-1))^color(blue)(2)))))/(0.00560^color(red)(2)color(darkgrey)(cancel(color(black)(("mol L"^(-1))^color(red)(2)))) * "0.00210 mol L"^(-1))# #K_c = "3887.3 mol"^(-1)"L"# Now, the equilibrium constant is usually given without added units, which means that your answer will be #K_c = color(green)(|bar(ul(color(white)(a/a)color(black)(3890)color(white)(a/a)|))) -># rounded to three sig figs Mind you, you can also have #color(blue)(2)"SO"_ (3(g)) rightleftharpoons color(red)(2)"SO"_ (2(g)) + "O"_ (2(g))# In this case, the equilibrium constant is #K_"c rev" = (["SO"_2]^color(red)(2) * ["O"_2])/(["SO"_3]^color(blue)(3))# Plug in your values to find #K_"c rev" = (0.00560^color(red)(2)color(darkgrey)(cancel(color(black)(("mol L"^(-1))^color(red)(2)))) * "0.00210 mol L"^(-1))/( 0.0160^color(blue)(2) color(darkgrey)(cancel(color(black)(("mol L"^(-1))^color(blue)(2)))))# #K_"c rev" = "0.000257 mol L"^(-1)# The answer would thus be #K_"c rev" = color(green)(|bar(ul(color(white)(a/a)color(black)(0.000257)color(white)(a/a)|))) -># rounded to three sig figs As a final note, notice that you have #color(purple)(|bar(ul(color(white)(a/a)color(black)(K_"c rev" = 1/K_c)color(white)(a/a)|)))#

Serenity Allen · Answer

undefined To calculate the equilibrium constant (K) for the reaction $2SO_2 + O_2 \rightleftharpoons 2SO_3$, use the expression for K:

$$ K = \frac{{[SO_3]^2}}{{[SO_2]^2 \times [O_2]}} $$

Substitute the given concentrations into the expression and calculate the value of K.

$$ K = \frac{{(0.0160 \, \text{mol/L})^2}}{{(0.00560 \, \text{mol/L})^2 \times (0.00210 \, \text{mol/L})}} $$

$$ K = \frac{{0.000256 \, \text{mol}^2/\text{L}^2}}{{0.00003136 \, \text{mol}^2/\text{L}^2}} $$

$$ K ≈ 8.16 $$

Therefore, the equilibrium constant (K) for the given reaction is approximately 8.16.