The reaction below has an equilibrium constant of K p=2.26×104 at 298 K . CO( g

Question

The reaction below has an equilibrium constant of
Kp=2.26×104 at 298 K.
CO(g)+2H2(g)CH3OH(g)

Part A

Calculate Kp for the reaction below.
12CH3OH(g)12CO(g)+H2(g)

SubmitMy AnswersGive Up

Part B

Predict whether reactants or products will be favored at equilibrium in the reaction above.

SubmitMy AnswersGive Up

Part C

Calculate Kp for the reaction below.
2CO(g)+4H2(g)2CH3OH(g)

SubmitMy AnswersGive Up

Part D

Predict whether reactants or products will be favored at equilibrium in the reaction above.

SubmitMy AnswersGive Up

Part E

Calculate Kp for the reaction below.
2CH3OH(g)2CO(g)+4H2(g)

SubmitMy AnswersGive Up

Part F

Predict whether reactants or products will be favored at equilibrium in the reaction above.

SubmitMy AnswersGive Up

The reaction below has an equilibrium constant of
Kp=2.26×104 at 298 K.
CO(g)+2H2(g)CH3OH(g)

Part A

Calculate Kp for the reaction below.
12CH3OH(g)12CO(g)+H2(g)

K =

SubmitMy AnswersGive Up

Part B

Predict whether reactants or products will be favored at equilibrium in the reaction above.

Predict whether reactants or products will be favored at equilibrium in the reaction above. Reactants will be favored at equilibrium. Products will be favored at equilibrium.

SubmitMy AnswersGive Up

Part C

Calculate Kp for the reaction below.
2CO(g)+4H2(g)2CH3OH(g)

K =

SubmitMy AnswersGive Up

Part D

Predict whether reactants or products will be favored at equilibrium in the reaction above.

Predict whether reactants or products will be favored at equilibrium in the reaction above. Reactants will be favored at equilibrium. Products will be favored at equilibrium.

SubmitMy AnswersGive Up

Part E

Calculate Kp for the reaction below.
2CH3OH(g)2CO(g)+4H2(g)

K =

SubmitMy AnswersGive Up

Part F

Predict whether reactants or products will be favored at equilibrium in the reaction above.

Predict whether reactants or products will be favored at equilibrium in the reaction above. Reactants will be favored at equilibrium. Products will be favored at equilibrium.

SubmitMy AnswersGive Up

Explanation / Answer

Kp=2.26×10^4

CO(g)+2H2(g) <------------------> CH3OH(g)

Kp1 = [CH3OH]/[CO][H2]^2

12CH3OH(g) <----------------------> 12CO(g)+24 H2(g)

Kp = [CO]^12 [H2]^24 / [CH3OH]^12

      = (1/Kp1)^12

     = (1/ 2.26×10^4 )^12

    = 5.63 x 10^-53

part B)

Reactants will be favored at equilibrium.

part C)

2CO(g)+4H2(g) <-------------------> 2CH3OH(g)

Kp = [CH3OH]^2 / [CO]^2 [H2]^4

Kp = (2.26 x 10^4)^2

Kp = 5.11 x 10^8

part D )

Products will be favored at equilibrium.

part E)

2CH3OH(g) <------------------> 2CO(g)+4H2(g)

part D inverse will give this

Kp = 1.96 x 10^-9

part F)

Reactants will be favored at equilibrium

note : Kp value is more reaction towards products side. Kp is less reaction towards reactant side

   

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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