At a certain temperature, the equilibrium constant K_c is 0.154 for the reaction

Question

At a certain temperature, the equilibrium constant K_c is 0.154 for the reaction 2SO_2(g) + O_2(g) doubleheadarrow 2SO_3(g) What concentration of SO_3 would be in equilibrium with 0.250 moles of SO_2 and 0.701 moles of O_2 in a 1.00 liter container at this temperature? A mixture consisting of 1 mol of H_2O(g) and 1 mol CO(g) is placed in a 14 L reaction vessel at 800 K. At equilibrium, 0.874 mol CO_2(g) is present as a result of the reaction CO(g) + H_2O(g) doubleheadarrow CO_2(g) + H_2(g). What is k_c at 800 K?

Explanation / Answer

004

Kc = [SO3]² / ([SO2]²* [O2])
0.154 = [SO3]² / {(0.250)² * 0.783}
0.154 = [SO3]² / 0.0489
[SO3]² = 0.154*0.0489
[SO3]² = 7.5*10^-3
[SO3] = 0.087 mol

005

Initially you have 1 mole of H2O and CO, and 0 moles of CO2 and H2.
At equilibrium, 0.374 moles of CO2 formed, so 0.374 moles of H2 would have formed also.

If 0.374 moles formed, then 0.374moles of H2O and CO must have reacted.
To find the moles at equilibrium, subtract initial moles by moles reacted.

Initially: 1 mol H2O
Amount reacted: 0.374 mol
Equilibrium moles: 1 - 0.374 = 0.626 mol

Same goes for CO.

Plug the equilibrium moles into the Kc formula:

[Products] / [Reactants]

( 0.374 *0.374) / (0.626 * 0.626) = 0.356

Kc = 0.356

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