Home/ Questions/The reaction below has an equilibrium constant of Kp=2.26×104 at 298 K.
Next
Answered
qhtutoria
  • 7
Asked: 2022-04-11T18:37:34+00:00 In:

The reaction below has an equilibrium constant of Kp=2.26×104 at 298 K.

  • 7

. Advertisement .

..3..

. Advertisement .

..4..

The reaction below has an equilibrium constant of Kp = 2.26×104 at 298 K.

CO(g) + 2H2(g) ⇌ CH3OH(g)

......... ADVERTISEMENT .........

..8..

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

Part A: Calculate Kp for the reaction below.
CH3OH(g) ⇌ CO(g)+2H2(g)
 
Part B: Reactants will be favored at equilibrium. O Products will be favored at equilibrium.
 
Part C: Calculate Kp for the reaction below. 
CO (g) + 2H2 (g) ⇌ CH3OH (g)
 
♦ Relevant knowledge
Equilibrium Constant

A reaction’s equilibrium constant expression of a reaction could be written using understanding that the expression of equilibrium for the reverse reaction, and is used to create an mathematical connection between them.

equilibriumkpreaction
Share

1 Answer

  1. Best Answer

    Given Data: The equilibrium constant is 2.26×104.

    A)

    The reaction described is illustrated below.

    CO(g) + 2H2(g) ↔ CH3OH(g)

    The formula to calculate equilibrium constant is provided below.

    Kp = [CH3OH] / ([CO][H2]2)⋅⋅⋅⋅⋅⋅(I)

    The value of the equilibrium constant is 2.26×104.

    B)

    The reaction that is described below.

    CH3OH(g) ↔ CO(g) + 2H2(g)

    The expression to calculate equilibrium constant is given below.

    K′p = ([CO][H2]2) / [CH3OH]⋅⋅⋅⋅⋅⋅(II)

    From equation (I) and (II),

    K′p = 1/Kp

    Substitute the value in the above expression.

    K′p = 1/(2.26×104) = 4.424×10−5

    So, the value calculated of the equilibrium constant is 4.424×10−5.

    C)

    Based on the calculations above It is evident from the above calculations that the equilibrium constant of the forward reaction will be considerably greater than that of the reverse reaction.

    Kp >>>> K′p

    Greater the value of equilibrium constant, more favorable is the reaction.

    Thus, the formation of products is favored during the reaction.

    • 21

Leave an answer

You must login to add an answer.

Need An Account,