The equilibrium constant K_c for the reaction is 0.534 at 700 degree C. H_2(g) +

Question

The equilibrium constant K_c for the reaction is 0.534 at 700 degree C. H_2(g) + CO_2(g) H_2O(g) + CO(g) Calculate the molarity of all reactants and products at equilibrium if a mixture of 0.300 m of CO_2, 0.300 mole of H_2,0.300 mole of H2O and 0.300 mole of CO is heated to 700 degree C 10.0-L container. Consider the heterogeneous equilibrium process: At 700 degree C, the total pressure of the system is found to be 4.50 atm. If the equilibrium K_p is 1.52, calculate the equilibrium partial pressures of CO_2 and CO.

Explanation / Answer

16.

     Reaction : H2(g) + CO2(g) <==> H2O(g) + CO(g)

Kc = [H2O][CO]/[H2][CO2]

initial concentration = [H2] = [CO2] = [H2O] = [CO] = 0.3/10 = 0.03 M

let x be the change in concentration of reactant at equilibrium then,

0.534 = (0.03+x)(0.03+x)/(0.03-x)(0.03-x)

0.534 = 9 x 10^-4 + 0.06x + x^2/9 x 10^-4 - 0.06x + x^2

4.806 x 10^-4 - 0.03204x + 0.534x^2 = 9 x 10^-4 + 0.06x + x^2

0.466x^2 + 0.3804x + 4.194 x 10^-4 = 0

x = -0.0011 M

Equilibrium concentration,

[H2] = 0.029 M

[CO2] = 0.029 M

[H2O] = 0.031 M

[CO] = 0.031 M

17.

Kp = [CO]^2/[CO2] = 1.52

[CO] + [CO2] = 4.50 atm

[CO2] = 4.50 - [CO]

1.52 = [CO]^2/(4.5 - [CO])

6.84 - 1.52[CO] = [CO]^2

[CO]^2 + 1.52[CO] - 6.84 = 0

So partial pressure of,

[CO] = 3.93 atm

[CO2] = 0.57 atm

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.