IB DP Chemistry Reactivity 1.4 - Entropy and spontaneity HL Paper 2- Exam Style Questions - New Syllabus
Question
Sulfur trioxide is an important compound in industry.
(a) Sulfur trioxide has more than one possible Lewis (electron dot) structure.
(i) Sketch two Lewis (electron dot) structures for SO3, one of which obeys the octet rule and one of which does not. [2]
(ii) State how chemists decide which Lewis (electron dot) structure is more stable. [1]
(iii) Predict the length of each S to O bond in pm. Data booklet (Section 10, covalent bond lengths): S–O (single) = 161 pm; S=S (double) = 189 pm; O=O (double) = 121 pm. [1]
(b) Suggest why atmospheric SO3(g) is an environmental concern. [1]
(c) State the name of a post-combustion method used to lower the quantity of SO3(g) released to the atmosphere. [1]
(d) SO3(g) is made using the contact process.
\[ \mathrm{2\,SO_2(g) + O_2(g) \rightleftharpoons 2\,SO_3(g)} \qquad \Delta H^\circ < 0 \]
(i) Sketch a potential energy profile for this reaction on the axes provided. Label \(E_\mathrm{a}\) and include labels on the axes. [3]
(ii) Explain why increasing the temperature increases the rate of reaction. [2]
(iii) Vanadium pentoxide, V2O5, is used as a catalyst. Explain how a catalyst increases the rate of a reaction. [2]
(iv) During the reaction, V2O5 changes to V2O4. Identify the oxidation states of vanadium in each compound. [1]
(v) State the equilibrium constant expression, \(K_\mathrm{c}\), for the production of 1 mol of SO3.
\[ \mathrm{SO_2(g) + \tfrac{1}{2}\,O_2(g) \rightleftharpoons SO_3(g)} \] [1]
\[ \mathrm{SO_2(g) + \tfrac{1}{2}\,O_2(g) \rightleftharpoons SO_3(g)} \] [1]
(vi) Calculate the entropy change, \(\Delta S^\circ\), in J K\(^{-1}\) mol\(^{-1}\), for the production of 1 mol of SO3(g). Use the absolute entropy values given in the table. [1]
| Substance | \(S^\circ\) / J K\(^{-1}\) mol\(^{-1}\) |
|---|---|
| SO2(g) | 248.2 |
| O2(g) | 205.2 |
| SO3(g) | 256.8 |
(vii) Outline, with reference to the equation, why the sign for the entropy change obtained in part (vi) is expected. [1]
(viii) Calculate the value of Gibbs free energy, \(\Delta G^\circ\), of the reaction, in kJ mol\(^{-1}\), at 773 K. Data booklet: use \(R=8.31\ \mathrm{J\,K^{-1}\,mol^{-1}}\) and \(\Delta H^\circ=-98.5\ \mathrm{kJ\,mol^{-1}}\). If you did not obtain an answer for (d)(vi), use \(-100\ \mathrm{J\,K^{-1}\,mol^{-1}}\), although this is not the correct answer. [1]
(ix) Calculate the value of the equilibrium constant for the formation of SO3(g) at 773 K. Data booklet: \(\Delta G^\circ=-RT\ln K\) and \(R=8.31\ \mathrm{J\,K^{-1}\,mol^{-1}}\) ( \(1\ \mathrm{kJ}=10^3\ \mathrm{J}\)). If you did not obtain an answer to (d)(viii), use \(-25.0\ \mathrm{kJ\,mol^{-1}}\), although this is not the correct answer. [2]
(x) A flask contains 0.120 mol dm\(^{-3}\) SO2(g), 0.050 mol dm\(^{-3}\) O2(g) and 0.150 mol dm\(^{-3}\) SO3(g) at 773 K. Deduce whether the system is at equilibrium and in which direction the reaction will proceed spontaneously if not at equilibrium. [2]
▶️ Answer/Explanation
Markscheme (with detailed working)
(a)(i)
Any two correct resonance Lewis structures, e.g. one octet-obeying (expanded octet not allowed) and one with expanded octet on S. Award for correct bonding/charges/electron pairs.
Obeys octet rule: Three resonance structures with double bonds to two oxygen atoms and a single bond to the third (each O has 8 electrons).
Does not obey octet rule: One double-bonded oxygen and two dative bonds from sulfur (expanded octet, 10 electrons on S).
[2]
Any two correct resonance Lewis structures, e.g. one octet-obeying (expanded octet not allowed) and one with expanded octet on S. Award for correct bonding/charges/electron pairs.
Obeys octet rule: Three resonance structures with double bonds to two oxygen atoms and a single bond to the third (each O has 8 electrons).
Does not obey octet rule: One double-bonded oxygen and two dative bonds from sulfur (expanded octet, 10 electrons on S).
[2]
(a)(ii)
Compare formal charges: structures with the lowest sum of formal charges and negative charge on the more electronegative atom (O) are more stable.
[1]
Compare formal charges: structures with the lowest sum of formal charges and negative charge on the more electronegative atom (O) are more stable.
[1]
(a)(iii)
All S–O bonds in SO3 are equivalent (resonance), so the bond length is between a single S–O (161 pm) and a double bond to O. A reasonable prediction is \(\approx 145\pm 5\ \mathrm{pm}\) (between 161 pm and the shorter double-bond benchmark such as O=O at 121 pm).
[1]
All S–O bonds in SO3 are equivalent (resonance), so the bond length is between a single S–O (161 pm) and a double bond to O. A reasonable prediction is \(\approx 145\pm 5\ \mathrm{pm}\) (between 161 pm and the shorter double-bond benchmark such as O=O at 121 pm).
[1]
(b)
SO3(g) forms H2SO4 (acid rain) in the atmosphere; contributes to aerosols/respiratory irritation/corrosion. Any one valid reason.
[1]
SO3(g) forms H2SO4 (acid rain) in the atmosphere; contributes to aerosols/respiratory irritation/corrosion. Any one valid reason.
[1]
(c)
Post-combustion flue-gas desulfurization (FGD), e.g. wet scrubbing with CaCO3/Ca(OH)2 to form CaSO4.
[1]
Post-combustion flue-gas desulfurization (FGD), e.g. wet scrubbing with CaCO3/Ca(OH)2 to form CaSO4.
[1]
(d)(i)
Potential energy profile: y-axis “Potential energy”, x-axis “Reaction progress”; reactants above products (exothermic). Activation energy \(E_\mathrm{a}\) shown from reactants to transition state. Include \(\Delta H^\circ\) drop to products. Correctly labelled and shaped curve.
[3]
Potential energy profile: y-axis “Potential energy”, x-axis “Reaction progress”; reactants above products (exothermic). Activation energy \(E_\mathrm{a}\) shown from reactants to transition state. Include \(\Delta H^\circ\) drop to products. Correctly labelled and shaped curve.
[3]
(d)(ii)
Higher \(T\) increases fraction of molecules with \(E \ge E_\mathrm{a}\) (Maxwell–Boltzmann distribution shifts), so more effective collisions per unit time; rate increases.
[2]
(d)(iii)
A catalyst provides an alternative pathway with lower \(E_\mathrm{a}\) (and/or surface for adsorption/desorption), increasing the rate without being consumed.
[2]
A catalyst provides an alternative pathway with lower \(E_\mathrm{a}\) (and/or surface for adsorption/desorption), increasing the rate without being consumed.
[2]
(d)(iv)
V in V2O5: +5; V in V2O4: +4.
[1]
V in V2O5: +5; V in V2O4: +4.
[1]
(d)(v)
\[ K_\mathrm{c} \;=\; \frac{[\mathrm{SO_3}]}{[\mathrm{SO_2}]\,[\mathrm{O_2}]^{1/2}}. \]
[1]
\[ K_\mathrm{c} \;=\; \frac{[\mathrm{SO_3}]}{[\mathrm{SO_2}]\,[\mathrm{O_2}]^{1/2}}. \]
[1]
(d)(vi)
For \(\mathrm{SO_2 + \tfrac{1}{2}O_2 \to SO_3}\):
\[ \Delta S^\circ \;=\; S^\circ(\mathrm{SO_3}) – \big(S^\circ(\mathrm{SO_2}) + \tfrac{1}{2}S^\circ(\mathrm{O_2})\big) \] \[ = 256.8 – \big(248.2 + \tfrac{1}{2}\times 205.2\big) \;=\; 256.8 – 350.8 \;=\; \boxed{-94.0\ \mathrm{J\,K^{-1}\,mol^{-1}}}. \]
[1]
For \(\mathrm{SO_2 + \tfrac{1}{2}O_2 \to SO_3}\):
\[ \Delta S^\circ \;=\; S^\circ(\mathrm{SO_3}) – \big(S^\circ(\mathrm{SO_2}) + \tfrac{1}{2}S^\circ(\mathrm{O_2})\big) \] \[ = 256.8 – \big(248.2 + \tfrac{1}{2}\times 205.2\big) \;=\; 256.8 – 350.8 \;=\; \boxed{-94.0\ \mathrm{J\,K^{-1}\,mol^{-1}}}. \]
[1]
(d)(vii)
Fewer moles of gas on the product side (1 mol) than reactant side (1.5 mol) → decreased disorder → \(\Delta S^\circ < 0\).
[1]
Fewer moles of gas on the product side (1 mol) than reactant side (1.5 mol) → decreased disorder → \(\Delta S^\circ < 0\).
[1]
(d)(viii)
\(\Delta G^\circ = \Delta H^\circ – T\Delta S^\circ\) with \(T=773\ \mathrm{K}\), \(\Delta H^\circ=-98.5\ \mathrm{kJ\,mol^{-1}}\), \(\Delta S^\circ=-94.0\ \mathrm{J\,K^{-1}\,mol^{-1}}=-0.0940\ \mathrm{kJ\,K^{-1}\,mol^{-1}}\):
\[ T\Delta S^\circ = 773 \times (-0.0940) = -72.662\ \mathrm{kJ\,mol^{-1}} \] \[ \Delta G^\circ = -98.5 – (-72.662) = \boxed{-25.84\ \mathrm{kJ\,mol^{-1}}\ (\approx -25.8)}. \]
[1]
\(\Delta G^\circ = \Delta H^\circ – T\Delta S^\circ\) with \(T=773\ \mathrm{K}\), \(\Delta H^\circ=-98.5\ \mathrm{kJ\,mol^{-1}}\), \(\Delta S^\circ=-94.0\ \mathrm{J\,K^{-1}\,mol^{-1}}=-0.0940\ \mathrm{kJ\,K^{-1}\,mol^{-1}}\):
\[ T\Delta S^\circ = 773 \times (-0.0940) = -72.662\ \mathrm{kJ\,mol^{-1}} \] \[ \Delta G^\circ = -98.5 – (-72.662) = \boxed{-25.84\ \mathrm{kJ\,mol^{-1}}\ (\approx -25.8)}. \]
[1]
(d)(ix)
\(\Delta G^\circ = -RT\ln K\) ⇒ \(K=\exp\!\big(-\Delta G^\circ/(RT)\big)\). Convert \(\Delta G^\circ\) to J mol\(^{-1}\): \(-25.84\ \mathrm{kJ\,mol^{-1}}=-2.584\times10^{4}\ \mathrm{J\,mol^{-1}}\). With \(R=8.31\ \mathrm{J\,K^{-1}\,mol^{-1}},\ T=773\ \mathrm{K}\):
\[ K = \exp\!\left(\frac{2.584\times10^{4}}{8.31\times 773}\right) = \exp(4.023) \approx \boxed{5.58\times 10^{1}}\ (\mathbf{56}). \] (Accept 50–60 depending on rounding; full credit for correct method.)
[2]
\(\Delta G^\circ = -RT\ln K\) ⇒ \(K=\exp\!\big(-\Delta G^\circ/(RT)\big)\). Convert \(\Delta G^\circ\) to J mol\(^{-1}\): \(-25.84\ \mathrm{kJ\,mol^{-1}}=-2.584\times10^{4}\ \mathrm{J\,mol^{-1}}\). With \(R=8.31\ \mathrm{J\,K^{-1}\,mol^{-1}},\ T=773\ \mathrm{K}\):
\[ K = \exp\!\left(\frac{2.584\times10^{4}}{8.31\times 773}\right) = \exp(4.023) \approx \boxed{5.58\times 10^{1}}\ (\mathbf{56}). \] (Accept 50–60 depending on rounding; full credit for correct method.)
[2]
(d)(x)
Reaction quotient: \[ Q_\mathrm{c}=\frac{[\mathrm{SO_3}]}{[\mathrm{SO_2}]\,[\mathrm{O_2}]^{1/2}} =\frac{0.150}{0.120\,(0.050)^{1/2}} =\frac{0.150}{0.120\times 0.2236} \approx \boxed{5.59}. \] Since \(Q_\mathrm{c} \lt K\) (\(\approx 56\)), the system is not at equilibrium and will proceed forward to form more SO3 spontaneously.
[2]
Reaction quotient: \[ Q_\mathrm{c}=\frac{[\mathrm{SO_3}]}{[\mathrm{SO_2}]\,[\mathrm{O_2}]^{1/2}} =\frac{0.150}{0.120\,(0.050)^{1/2}} =\frac{0.150}{0.120\times 0.2236} \approx \boxed{5.59}. \] Since \(Q_\mathrm{c} \lt K\) (\(\approx 56\)), the system is not at equilibrium and will proceed forward to form more SO3 spontaneously.
[2]
Total Marks: 22