A student ran the following reaction in the laboratory at 1089 K: 2SO3(g) 2SO2(g) + O2(g) When she introduced 8.39×10-2 moles of SO3(g) into a 1.00 liter container, she found the equilibrium concentration of O2(g) to be 1.78×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction.

The equilibrium constant, Kc is 0.00967

Explanation:

Step 1: Data given

Temperature = 1089 K

Number of moles SO3 = 8.39 * 10^-2 moles = 0.0839 moles

Volume = 1.0 L

The equilibrium concentration of O2(g) to be 1.78* 10^-2 M

Step 2: The balanced equation

2SO3(g) ⇆ 2SO2(g) + O2(g)

Step 3: The initial concentration

Concentration = moles / volume

[SO3] = 0.0839 moles / 1 L = 0.0839 M

[SO2] = 0M

[O2] = 0M

Step 4: The concentration at equilibrium

For 2 moles SO3 we'll have 2 moles SO2 and 1 mol O2

[SO3] = 0.0839 - 2X M

[SO2] = 2X M

[O2] = XM = 1.78 * 10^-2 M = 0.0178 M

So X = 0.0178

[SO3] = 0.0839 - 2*0.0178 = 0.0483 M

[SO2] = 2*0.0178 = 0.0356 M

[O2] = XM = 1.78 * 10^-2 M = 0.0178 M

Step 5: Calculate the equilibrium constant, Kc

Kc = [O2][SO2]² / [SO3]²

Kc = (0.0178 * 0.0356²)/0.0483²

Kc = 0.00967

The equilibrium constant, Kc is 0.00967

answer: h+ its less than oh

explanation:

i think it's c. i'm not sure but i think so.