CHEM1101 Worksheet 11 – Answers to Critical Thinking Questions The worksheets are available in the tutorials and form an integral part of the learning outcomes and experience for this unit. Model 1: The Equilibrium Constant 1. K c (A) = [! ! ! ! ! ] [!! ! ! ] ! K c (B) = [! ! ! ! ! ] !/! [!! ! ! ] K c (C) = [!! ! ! ] ! [! ! ! ! ! ] K c (D) = [!! ! ! ] [! ! ! ! ! ] !/! 2. (a) K c (B) = ! (A) (b) K c (A) = 1 / K c (C) 3. K c (A) = 0.078, K c (B) = 0.28, K c (C) = 13. Model 2: The Reaction Quotient 1. The reaction will shift to the right to decrease [NO 2 (g)]. 2. The reaction will shift to the left to increase [NO 2 (g)]. 3. (a) Q c = 0.050 (b) Q c = 0.20. 4. (a) If Q c < K c , the reaction will shift to the right. (b) If Q c > K c , the reaction will shift to the left. Model 3: Equilibrium calculations Critical thinking questions 1. See table opposite. 2. See table opposite. 3. K c (A) = [! ! ! ! ! ] [!! ! ! ] ! = (!.!"!!) (!.!!!!") ! 4. x = 0.070 M so [NO 2 (g)] = 1.86 M and [N 2 O 4 (g)] = 0.27 M (The second root is non-physical as it leads to a negative concentration for NO 2 . Model 4: Enthalpy (Δ rxn H) and Entropy (Δ rxn S) of Reaction 1. Δ rxn H° = -57 kJ mol -1 . Δ rxn S° = -176 J K -1 mol -1 2. The reaction involves making a N-N bond, with no bonds being broken. It is exothermic. The reaction involves the conversion of 2 mol of gas 1 mol of gas. The entropy decreases. 3. Δ rxn H° = -28.5 kJ mol -1 . Δ rxn S° = -88 J K -1 mol -1 . These values are exactly half those for reaction A. 4. Δ rxn H° = +57 kJ mol -1 . Δ rxn S° = +176 J K -1 mol -1 . These values are equal to -1 times the values for reaction A. Reaction C involves breaking a N-N bond, with no bonds being made. It is endothermic. The reaction involves the conversion of 1 mol of gas 2 mol of gas. The entropy increases.. 5. Δ rxn H° = +28.5 kJ mol -1 . Δ rxn S° = +88 J K -1 mol -1 . 2NO 2 (g) N 2 O 4 (g) initial 2.00 0.20 change -2x +x equilibrium 2.00 – 2x 0.20 + x
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CHEM1101 Worksheet 11 – Answers to Critical Thinking Questions The worksheets are available in the tutorials and form an integral part of the learning outcomes and experience for this unit.
Model 1: The Equilibrium Constant
1. Kc (A) = [!!!! ! ][!!! ! ]!
Kc (B) = [!!!! ! ]!/!
[!!! ! ]
Kc (C) = [!!! ! ]!
[!!!! ! ] Kc (D) =
[!!! ! ][!!!! ! ]!/!
2. (a) Kc (B) = 𝐾! (A) (b) Kc (A) = 1 / Kc (C)
3. Kc (A) = 0.078, Kc (B) = 0.28, Kc (C) = 13.
Model 2: The Reaction Quotient 1. The reaction will shift to the right to decrease [NO2(g)].
2. The reaction will shift to the left to increase [NO2(g)]. 3. (a) Qc = 0.050
(b) Qc = 0.20. 4. (a) If Qc < Kc, the reaction will shift to the right.
(b) If Qc > Kc, the reaction will shift to the left.
Model 3: Equilibrium calculations
Critical thinking questions 1. See table opposite.
2. See table opposite.
3. Kc (A) = [!!!! ! ][!!! ! ]!
= (!.!"!!)(!.!!!!")!
4. x = 0.070 M so [NO2(g)] = 1.86 M and [N2O4(g)] = 0.27 M
(The second root is non-physical as it leads to a negative concentration for NO2.
Model 4: Enthalpy (ΔrxnH) and Entropy (ΔrxnS) of Reaction 1. ΔrxnH° = -57 kJ mol-1. ΔrxnS° = -176 J K-1 mol-1
2. The reaction involves making a N-N bond, with no bonds being broken. It is exothermic.
The reaction involves the conversion of 2 mol of gas à 1 mol of gas. The entropy decreases.
3. ΔrxnH° = -28.5 kJ mol-1. ΔrxnS° = -88 J K-1 mol-1. These values are exactly half those for reaction A.
4. ΔrxnH° = +57 kJ mol-1. ΔrxnS° = +176 J K-1 mol-1. These values are equal to -1 times the values for reaction A. Reaction C involves breaking a N-N bond, with no bonds being made. It is endothermic. The reaction involves the conversion of 1 mol of gas à 2 mol of gas. The entropy increases..
5. ΔrxnH° = +28.5 kJ mol-1. ΔrxnS° = +88 J K-1 mol-1.
2NO2(g) N2O4(g)
initial 2.00 0.20
change -2x +x
equilibrium 2.00 – 2x 0.20 + x
CHEM1101 2013-J-9 June 2013
• Consider the following reaction.
N2O4(g) 2NO2(g)
An equilibrium mixture in a 1.00 L container is found to contain [N2O4] = 1.00 M and [NO2] = 0.46 M. The vessel is then compressed to half its original volume while the temperature is kept constant. Calculate the concentration [N2O4] when the compressed system has come to equilibrium. Show all working.
Marks 4
For this reaction, the equilibrium constant expression is given by:
Kc = [𝐍𝐎𝟐 𝐠 ]𝟐
[𝐍𝟐𝐎𝟒]
As mixture is at equilibrium when [N2O4] = 1.00 M and [NO2] = 0.46 M:
Kc = (𝟎.𝟒𝟔)𝟐
(𝟏.𝟎𝟎) = 0.21
If the volume of the vessel is halved, the initial concentrations will double: [N2O4] = 2.00 M and [NO2] = 0.92 M. The reaction is no longer at equilibrium and Le Chatelier’s principle predicts it will shift towards the side with fewer moles: it will shift towards reactants.
A reaction table needs to be used to calculate the new equilibrium concentrations.
N2O4(g) 2NO2(g) initial 2.00 0.92
change +x -2x
equilibrium 2.00 + x 0.92 – 2x
Hence,
Kc = [𝐍𝐎𝟐 𝐠 ]𝟐
[𝐍𝟐𝐎𝟒] = (𝟎.𝟗𝟐!𝟐𝒙)
𝟐
(𝟐.𝟎𝟎!𝒙) = 0.21
So,
(0.92 – 2x)2 = 0.21 (2.00 + x)
0.8464 - 3.68x + 4x2 = 0.42 + 0.21x
4x2 – 3.89x + 0.43 = 0
ANSWER CONTINUES ON THE NEXT PAGE
With a = 4, b = -3.89 and c = 0.43, this quadratic equation has roots:
x = !𝒃 ± 𝒃𝟐!𝟒𝒂𝒄
𝟐𝒂 = 𝟑.𝟖𝟗 ± (!𝟑.𝟖𝟗)𝟐!𝟒×𝟒×𝟎.𝟒𝟑
𝟐×𝟒
This gives x = 0.13 or 0.85. The latter makes no chemical sense as it gives a negative concentration for NO2.
Hence using x = 0.13:
[N2O4] = (2.00 + x) M = (2.00 + 0.13) M = 2.13 M
[NO2] = (0.92 - 2x) M = (0.92 – 2 × 0.13) M = 0.66 M
Answer: [N2O4] = 2.13 M
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