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
INTERACTIVE EXAMPLE Solving an Equilibrium Problem (Involving a Linear Equation in x) The reaction CO(g) +...
www .locatoriamente INTERACTIVE EXAMPLE Solving an Equilibrium Problem (Involving a Linear Equation in x) Chapter 54 Homework for Grade Question 1 1pl Question 2 1 pt Question 3 Question 4 pl The reaction CO(g) + H2O(g) + CO2(&)-H (8) has an equilibrium constant of 0.58 at 1000 °C. If 20.0-L mixture of CO and H.Ohas a concentration of each of 2.00x10"moll, what amount (in moles) of each species will be present when the mixture reaches equilibrium? mol CO Questions for...
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