The equilibrium constant, K c for the reaction of H2 and I2 is 54 at 425C. H2( g

Question

The equilibrium constant, Kc for the reaction of H2 and I2 is 54 at 425C.
H2(g)+I2(g)2HI(g)

If the equilibrium mixture contains 0.015 M I2 and 0.035 M HI, what is the concentration of H2?

Express your answer using two significant figures.

[H2] =             M

Indicate any changes that a - d will cause for the following reaction initially at equilibrium:
C2H4(g)+Cl2(g)C2H4Cl2(g)+heat

Part A

raise the temperature of the reaction

C) The equilibrium position does not change.

Part B

decrease the volume of the reaction container

Part C

add a catalyst

Part D

add more Cl2(g)

A) The system shifts in the direction of the reactants. B) The system shifts in the direction of the products.

C) The equilibrium position does not change.

Part B

decrease the volume of the reaction container

A)The system shifts in the direction of the reactants. B)The system shifts in the direction of the products. C) The equilibrium position does not change.

Part C

add a catalyst

A) The system shifts in the direction of the reactants. B) The system shifts in the direction of the products. C) The equilibrium position does not change.

Part D

add more Cl2(g)

A) The system shifts in the direction of the reactants. B) The system shifts in the direction of the products. C) The equilibrium position does not change.

Explanation / Answer

1)

H2 + I2 <-> 2HI

K = [HI]^2 / ([H2][I2])

K = (0.035)^/(0.015)/(H2]

54 = 0.0816666/[H2]

[H2] = 0.0816666/54 = 0.001512

2)

C2H4(g)+Cl2(g)C2H4Cl2(g)+heat

this is clearly exothermic, since it releases heat

If T increases, then,equilibrium shifts to left (reactants)

B)

decrease in V, means increase in P, less moles are favoured

Then products are favoured since n = 1

C)

adding a catlayst wont chang eequilibrium conditions

D)

adding Cl2, will make a shift twards right, the products formation

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