For the reaction below predict the effect of each of the following changes on th

ID: 482168 • Letter: F

Question

For the reaction below predict the effect of each of the following changes on the equilibrium position. (For each change put an X in the correct column) H_2 (g) + I_2 (g) doubleheadarrow 2HI (g) Delta H = -150 kJ/mol b. Consider the following reaction at 600K: H_2 (g) I_2 doubleheadarrow (g) 2HI (g) A reaction mixture at equilibrium at 600K contains 0.170 bar H_2, 0.120 bar I_2 and 1.26 bar HI. a. Calculate the equilibrium constant (K_p) for this reaction b. A second reaction mixture, also at 600 K contains 0.135 bar of H_2, 0.135 bar I_2 and 1.10 bar of HI. Is the mixture at equilibrium? If the mixture in part b is not at equilibrium, what will be the partial pressures of H_2, I_2 and HI when the reaction mixture reaches equilibrium? d. What will be the total pressure in the flask at equilibrium?

Explanation / Answer

H2 (g) + I2 (g) 2 HI (g) H = -150 kJmol-1

Change

Shift towards product

Shift towards reactant

No change

Removal of H2

Decreasing the temperature

Decreasing the container volume

Addition of I2

Addition of inert gas such as He

i) On removal of H2, the equilibrium of the reaction is disturbed. In order to restore it, the reaction proceeds in a direction of reactant.

ii) The equilibrium constant for an exothermic reaction (- H) increases as the temperature decreases Position of equilibrium will favor the forward direction (right).

iii) No change to the position of equilibrium, both sides have the same number of moles of gas.

iv)Increase in concentration of a reactant I2, Position of equilibrium will favor the forward direction (right)

v) If the volume is kept constant and an inert gas such as He added which does not take part in the reaction, the equilibrium remains undisturbed.

b) a.H2 (g) + I2 (g) 2 HI 600K

0.170 bar H2

0.120 bar I2

1.26 bar HI

KP = P2 HI/P H2 x P I2

KP = (1.26)^2/0.170 x 0.120 = 77.82

b. H2 (g) + I2 (g) --à 2 HI 600K

0.135 bar H2

0.135 bar I2

1.10 bar HI

KP = P2 HI/P H2 x P I2

KP = (1.10)^2/(0.135 x 0.135) = 66.39

d) P total = 0.170 bar H2 + 0.120 bar I2 +1.26 bar HI = 1.55bar