CO2(g) + H2(g)% H2O(g) + CO(g) When H/(g) is mixed with CO-(g) at 2,000 K, equil

Question

CO2(g) + H2(g)% H2O(g) + CO(g) When H/(g) is mixed with CO-(g) at 2,000 K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured [HJ = 0.20 mol/L [CO,] = 0.30 mol/L [Hol [CO-0.55 mol/L What is the mole fraction of CO(g) in the equilibrium mixture? (a) (b) Using the equilibrium concentrations given above, calculate the value of K, the equilibrium constant for the reaction. (c) Determine Kp in terms of Ke for this system. (d) When the system is cooled from 2,000 K to a lower temperature, 30.0 percent of the CO(g) is converted back to CO4g) Calculate the value of Ke at this lower temperature. (e) In a different experiment, 0.50 mole of H(g) is mixed with 0.50 mole of CO(g) in a 3.0-liter reaction vessel at 2,000 K. Calculate the equilibrium concentration, in moles per liter, of COg) at this

Explanation / Answer

1)mole fraction=(0.55/0.55+0.55+0.2+0.3)=0.343

2)Kc=(0.55X0.55)/(0.2X0.3)=5.04

3)Kp=Kc(RT)^delta n=5.04 as deita n=0

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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