Question

INTERACTIVE EXAMPLE Solving an Equilibrium Problem (Involving a Linear Equation in x) The reaction CO(g) + H2O(g) + CO2(g) + H2(g) has an equilibrium constant K of 0.58 at 1000 °C. If a 20.0-L mixture of CO and H2O has a concentration of each of 6,00x10* moll, what amount (in moles) of each species will be present when the mixture reaches equilibrium? mol CO mol H20 mol CO2 mol H2 Submit

Answer 1

The reaction with initial and equilibrium concentrations is:

CO (g) + H2O (g) ====== CO2 (g) + H2 (g)  

Initial 6.00*10-2M 6.00*10-2M 0 0

After x moles each of CO and H2O have reacted,

Equilibrium    (6.00*10-2 -x) (6.00*10-2 -x) x x

Now,

Kc can be written as:

Kc = ([ CO2] * [ H2 ]) / ([CO] * [ H2O]) which is,

Kc = (x * x ) /   ((6.00*10-2 -x) *   (6.00*10-2 -x) )

Kc = x2/   (6.00*10-2 -x) 2 , Given Kc = 0.58

So, x/  (6.00*10-2 -x) = (0.58)1/2

x/  (6.00*10-2 -x) = 0.761

On, solving this we get,

x = 0.026

Now, number of moles = Concentration * Volume

Given, volume = 20L

So, number of moles of CO2 = (x*20) = number of moles of H2 = 0.52 moles

number of moles of CO = (6.00*10-2- x) *20 = number of moles of H2O = (6.00*10-2 -0.026) * 20 = 0.68 moles