The equilibrium constant is 8.8 x 10 2 at a particular temperature for the react

Question

The equilibrium constant is 8.8 x 102 at a particular temperature for the reaction:

H2(g) + I2(g) 2HI(g)

Use this information to decide what will happen, given the following sets of initial conditions:
Will the reaction shift to the left, to the right or will the system be at equilibrium?


A 1.0 L flask contains 0.0454 mol H2, 0.011 mol I2 and 0.663 mol HI.

A 2.0 L flask contains 1.21 mol H2, 0.077 mol I2 and 8.41 mol HI.

PH2 = 0.35 torr, PI2 = 0.062 torr, PHI = 6.02 torr

PH2 = 0.5625 torr,PI2 = 0.062 torr, PHI = 5.54 torr

Explanation / Answer

The equilibrium constant is 8.8 x 102 at a particular temperature for the reaction:

Given chemical transformation is,

H2(g) + I2(g) 2HI(g)

Number of moles of gaseous product Np = 2

And Number of moles of gaseous reactant Nr = 1+1 = 2

n = Np-Nr = 2-2 = 0

Kp = Kc numerically same.

Kp = P2HI / PH2 x PI2

Kc = [HI]2/[H2][I2]

And given that the equilibrium constant

Kp = Kc =8.8 x 102

For any Kp (Or Kc) < 8.8 x 102 : reaction in reverse direction i.e. Reactant formation favored.

And for any Kp (or Kc) > 8.8 x 102 : reaction in forward direction i.e. Product formation favored.

For Kp (or Kc) = 8.8 x 102 : reaction is at equilibrium and favored in neither direction.

Based on this let us find appropriate equilibrium (Kp of Kc ) for given set of conditions,

1) A 1.0 L flask contains 0.0454 mol H2, 0.011 mol I2 and 0.663 mol HI.

[HI] = 0.663 moles /1 L = 0.663 M/L,

[I2] = 0.011moles /1L = 0.011 M/L,

[H2] = 0.0454 mol/ 1L = 0.0454 M/L

Using these concentrations, let us calculate new equilibrium constant say K1

K1 = [HI]2/[H2][I2]

K1 = (0.663)2/(0.0454)(0.011)

K1 = 8.02 x 102

K1 < 8.8 x 102

K1 < Kc

Hence reaction in reverse direction will be favored.

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2) A 2.0 L flask contains 1.21 mol H2, 0.077 mol I2 and 8.41 mol HI.

[HI] = 8.41 moles /2 L = 4.205 M/L,

[I2] = 0.077moles /2L = 0.0385 M/L,

[H2] = 1.21 mol/ 1L = 0.605 M/L

Using these concentrations, let us calculate new equilibrium constant say K1,

K2 = [HI]2/[H2][I2]

K2 = (4.205)2/(0.0385)(0.605)

K2 = 7.59 x 102

K2 < 8.8 x 102

K2 < Kc

Hence reaction in reverse direction will be favored.

3) PH2 = 0.35 torr, PI2 = 0.062 torr, PHI = 6.02 torr

Let new equilibrium constant be K3,

K3 = P2HI / PH2 x PI2

K3 = (6.02)2 /(0.062)(0.35)

K3 = 16.70 x 102

K3 > 8.8 x 102

K3 > Kp

Hence forward reaction i.e. product formation will be favored.

4) PH2 = 0.5625 torr,PI2 = 0.062 torr, PHI = 5.54 torr

Let new equilibrium constant be K4,

K4 = P2HI / PH2 x PI2

K4 = (5.54)2 /(0.062)(0.5625)

K4 = 16.70 x 102

K4 = 8.8 x 102

K4 = Kp

Hence reaction will be favored in neither direction.

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