Chemical Equilibrium Worksheet 1 Answers

Chemical Equilibrium Worksheet 1 (Suggested Answer) 1 a) Fe catalyst, 200 atm 450 oC b)

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N2 and H2 have strong bonds, hence high temperature is needed as Ea of reaction is high. (not elaborate explanation.)

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As high temperature favours the equilibrium position to shift to the left, favouring backward reaction.

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A catalyst is used such that not too high a temperature is used as it lowers Ea of the forward and backward reaction and increased the rate by the same extent, hence causing equilibrium position to remain the same but rate in which equilibrium reached is faster.

In this manner, yield and rate will not be compromised for one over the other. [Note: 3rd and 4th points are often missed out because students forget that catalyst allow a fast rate of reaction to occur at a lower temperature because the Ea is lowered.] 2 (i) PNH [ NH ] 2 ƒ

c)

Kp =

(ii)

3

PH3 2 PN 2

Kc =

3

[ N 2 ][ H 2 ] 3

You can assume that H2 and N2 is in the mole ratio of 3:1 Initial/ mole Change/ mole Equilibrium/ mole

3H2 (g) 3 -0.9 2.1

+ N2 (g) 1 -0.3 0.7

2NH3 (g) 0 +0.6 0.6

PH 2 = (2.1/3.4) x 190 = 117.3 atm PNH 3 = (0.6/3.4) x 190 = 33.52 atm PN 2 = (0.7/3.4) x 190 = 39.11 atm Kp =

33.52 2 3

= 1.78 × 10 −5 atm − 2

(117.3) (39.11)

(Note: If you assumed H2 and N2 were in equimolar, it is conceptually correct if you have done by the same steps) (iii) Since PV = nRT, P x (1.01 x 105) = RT(n/(Vx103)) = RT([substance]) X 10-3 P = RT([substance]) x 10-3 / (1.01 x 105) = [(RT/1.01) x [Subst] x 10-8 Kc = (RT/1.01x108)2Kp = 5.46 x 10-14 mol-2 dm6 [Mathematically intensive. P = Nm-2, V = m3 (which you will need to convert)]

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Chemical Equilibrium

d) (i)

ƒ By LCP, when pressure increase, the equilibrium position shifts to the right. ƒ Forward reaction is favoured because it produces less moles of gases (From 4 to 2) => state the number. ƒ It is the equilibrium’s response to increase pressure, hence in tries to reduce it. ƒ At new equilibrium, n(NH3) increases, while n(H2) and n(N2) decreases.

(ii)

ƒ There is no change to the equilibrium position. ƒ Although adding an inert gas causes the total pressure to increase, the individual partial pressures of N2, H2 and NH3 at equilibrium remains the same.

e) (i)

ƒ As strong N2 bonds are broken. Hence, Ea is high. Therefore, high temperature is required. ƒ In addition, high temperature favours the equilibrium position to shift to the right, favouring the forward reaction. Thus, increasing yield of NO. ƒ A catalyst is not needed as high temperature favours rate and yield. ƒ There is no need for a catalyst to lower the temperature of the reaction such that the rate of the reaction is compromise for the yield.

(ii)

Bond broken: 1 x N≡N and 1 O = O. Bond formed: 2 bonds between N and O. Bond energy of bond between N and O = ½(994 + 496 – (+90.2)) = 700 kJmol-1

f)

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Dissolving of non-metal oxide into water is to form an acidic solution. It is not a redox reaction.

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NO’s N oxidation number is + 2

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HNO3’s N oxidation number is + 5

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Hence, it is not possible to dissolve NO to give HNO3 as that would mean that H2O oxidizes NO.

ƒ (Water is unable to oxidize non-metal oxides, usually only O2 can) [Note: students thought that the question was asking whether NO was soluble. But if that is the intension of the question, then why would it tell you about HNO3?]

Kwok YL

Chemical Equilibrium