Question

References Use the References to access important values if needed for this question A student ran the following reaction in the laboratory at 1185 K: 2sO (g)+O2(8)2s05(8) When she introduced 8.19x102 moles of So,(g) and 8.10x10 moles of O2(g) into a 1.00 liter container, she found the equilibrium concentration of O2(g) to be 5.98x102 M. Calculate the equilibrium constant, Ke she obtained for this reaction. Ke Submit Answer

Answer 1

Answer:

19.27

Explanation:

Some values are corrected from correct source.Thus,

Moles of SO₂ = 8.19x10⁻² moles

Moles of O₂ = 8.10x10⁻² moles

Volume = 1 L

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Concentration of SO₂ = 8.19x10² M

Concentration of O₂ = 8.10x10 M

Considering the ICE table for the equilibrium as:

                    2SO₂ (g) +    O₂ (g)     ⇔          2SO₃ (g)

t = o              8.19x10⁻²      8.19x10⁻²

t = eq                -2x                  -x                      2x

--------------------------------------------- --------------------------

neq:    8.19x10⁻² -2x       8.19x10⁻² -x                      2x

Given:  

Equilibrium concentration of  O₂ = 5.98x10⁻² M  = 8.19x10⁻² -x

Thus, x = 0.0212 M

[SO₂] = 8.19x10⁻² - 2*0.0212 = 0.0395 M

[SO₃] = 2*0.0212 = 0.0424 M

The expression for the equilibrium constant is:

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

$K_c=\frac{{0.0424}^2}{{0.0395}^2\times 0.0598}$  

K = 19.27