At a certain temperature, 0.660 mol of SO3 is placed in a 4.50-L container, according to the reaction: 2SO3(g) --> 2SO2(g) + O2(g), at equilibrium, 0.130 mol of O2 is present, how would you calculate Kc?

Answer 1

#K_c = 0.0122#

The equilibrium constant for a given equilibrium reaction can be found by raising each equilibrium concentration to the power of its corresponding stoichiometric coefficient and dividing the result by the equilibrium concentrations of the reactants and products.

Regarding your balanced response

#color(red)(2)"SO"_text(3(g]) rightleftharpoons color(blue)(2)"SO"_text(2(g]) + "O"_text(2(g])#
the equilibrium constant, #K_c#, will take the form
#K_c = ( ["SO"_2]^color(blue)(2) * ["O"_2])/(["SO"_3]^color(red)(2))#

You are aware that the initial sulfur trioxide concentration is equivalent to

#color(blue)(c = n/V)#
#["SO"_3]_0 = "0.660 moles"/"4.50 L" = "0.1467 M"#

Since there is only the reactant in the reaction container at first, the starting concentrations of the two products will be equal to zero.

This means that once the reaction begins, you can use an ICE table to help you figure out how the species' concentrations will change.

#" " color(red)(2)"SO"_text(3(g]) " "rightleftharpoons" " color(blue)(2)"SO"_text(2(g]) " "+" " "O"_text(2(g])#
#color(purple)("I")" " " "0.1467" " " " " " " " " "0" " " " " " " " " "0# #color(purple)("C")" " " "(-color(red)(2)x)" " " " " "(+color(blue)(2)x)" " " " " "(+x)# #color(purple)("E")" "0.1467-color(red)(2)x" " " " " "color(blue)(2)x" " " " " " " " " " "x#
Now, you know that the reaction container will contain #0.130# moles of oxygen gas at equilibrium. This means that the equilibrium concentration of oxygen gas will be equal to
#["O"_2] = "0.130 moles"/"4.50 L" = "0.02889 M"#
Now, take a look at the ICE table. The equilibrium concentration of oxygen gas is said to be equal to #x#. This means that the equilibrium concentrations of the other two chemical species involved in the reaction will be
#["SO"_3] = 0.1467 - 2 * 0.02889 = "0.08892 M"#
#["SO"_2] = 2 * 0.02889 = "0.05778 M"#

This indicates that, for this reaction, the equilibrium constant at this particular temperature will be

#K_c = ( 0.05778^2 * 0.02889)/0.08892^2 = color(green)(0.0122)#

Three sig figs are used to round the result.

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Answer 2

To calculate the equilibrium constant ( K_c ), you use the concentrations of the reactants and products at equilibrium raised to the power of their respective stoichiometric coefficients. In this case:

[ K_c = \frac{[SO_2]^2 \cdot [O_2]}{[SO_3]^2} ]

Given that ( [SO_3] = \frac{0.660 \text{ mol}}{4.50 \text{ L}} ), ( [O_2] = \frac{0.130 \text{ mol}}{4.50 \text{ L}} ), and ( [SO_2] = \frac{2 \times 0.660 \text{ mol}}{4.50 \text{ L}} ), you can substitute these values into the equation to calculate ( K_c ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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