Chemical Equilibrium (Worksheet 2 / Suggested Answer) 1 a) (i) When a reversible reaction whose forward reaction and backward reaction

have the same rate.

(ii)

0 Timet

At time t, the equilibrium is attained.

Graph of change of rate of forward reaction with time.

Graph of change of rate of backward reaction with time.

Rates of reaction

(Note: Please ensure that the origin is there and the graph clearly labeled.)

(iii) The equilibrium cannot be achieved if a sealed vessel is not used. This is because when HI decomposes, H2 and I2 can escape. This causes the equilibrium position to constantly shift right, until all the HI is decomposed

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

Initial/mole: x 0 0 Change/mole -0.25x +0.125x -0.125x Equilibrium/mole: 0.75x 0.125x 0.125x Mole fraction of HI = 0.75x/ (0.75x + 0.125x + 0.125x) = 0.75 Mole fraction of H2 = 0.125 x / (x) = 0.125 = mole fraction of I2 Partial pressure of HI = 0.75 x 1.6 = 1.2 atm Partial pressure of H2 = 0.125 x 1.6 = 0.2 atm Partial pressure of I2 = 0.2 atm

0277.02.1

)2.0(2

2

222 ===

HI

IHp P

PPK

(Note: The 1.6 atm refers to equilibrium pressure. This is because it is only at equilibrium then we do know that 25% is decomposed. Hence, 300 oC and 1.6 atm refers to equilibrium conditions.)

(ii) 0.0711 (Kp = Kc)

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

Initial/atm: 1.2 0.4 0 Change/atm 1.2 – 2x 0.4 + x x Equilibrium/atm: 1.2 – 2x 0.4 + x x

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02777.0)22.1(

))(4.0(22

22 =−+

==x

xxP

PPK

HI

IHp

0.4x + x2 = (0.02777) (1.44 – 4.8 x + 4x2) 0.8889 x2 + 0.5333x – 0.0399= 0

0672.0667.0)8889.0(26529.05333.0

)8889.0(26529.05333.0

)8889.0(2)0399.0)(8889.0(45333.05333.0 2

orx

or

x

−=

+−−−=

−−±−=

x = 0.0672 Since V is constant, P α n Percentage decomposition = [(0.1169 x 2) / 1.2 ] x 100% = 11.2% (Note: The 1.2 and 0.4 atm in the question does not refer to equilibrium partial pressure. This is because it will imply that the total pressure will be greater than 1.6 atm (as H2 must have a partial pressure). This implies that n will have to increase for P to increase, since T and V are constants. But, the reaction does not increase number of gas particles. Therefore, the values in the question are not equilibrium partial pressure.)

(iv) The answer in (a)(iii) is smaller. Presence of H2 initially causes the equilibrium position to shift to left. This results in the backward reaction to be favoured. The extent of the forward reaction decreases.

(Note: Please use the structure. It is important to relate that because equilibrium position has shift to the left, the extent of the forward reaction decrease.)

c)

ttime

Concentration

HI

I2

(ii)(i)

0

12.5%

75%

100%

(Note: The initial increase in concentration is due to P ↑. But V ↓, hence concentration ↑. But equilibrium position does not change, hence n(gases) remains. Therefore we don’t see a gradual change (implies equilibrium’s response) in concentration. Sharp changes usually implies disturbance.) (Note: The 100%, 75% and 12.5% makes use of the Data’s information.)

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d) (i) ∆H of forward reaction = 2(299) – (436 + 151) = +11 kJmol-1

Increasing temperature. Equilibrium will shift to the right. Forward reaction is favoured because it is endothermic. This is equilibrium response to increasing temperature as it tries to

remove the heat from the increasing temperature. At new equilibrium the amount of H2 and I2 increase while HI decreases.

(ii) Forward reaction rate increases.

Backward reaction rate increases. K = rate constant of fwd/rate constant of backward. Forward reaction rate increases more; rate constant of fwd increase

more. K increases.

(Note: LCP is not applicable here as you told to use the definition of dynamic equilibrium.)

e) (i)

2H2 (g) + I2 (s)

H2 (g) + I2 (l) H2 (g) + I2 (g)ΔHvap (I2)

ΔHfus (I2)

ΔHf

ΔHr = -11 kJ mol-1

∆Hvap I2 = (-15.5) + 2(26.5) – (-11) = +48.5 kJ mol-1

(ii) ∆S = +(48.5 x 1000)/(184.4 + 273) = +106 J mol-1 K-1

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