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IB Chemistry R1.4: Entropy and spontaneity worked solutions

Worked solutions to every practice question in R1.4: Entropy and spontaneity of Photon Chemistry: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question B1

entropy-states[1 mark]
A gas in a rigid sealed vessel is cooled without condensation. Which explanation describes its entropy change?
A   It decreases because fewer energy distributions are accessible. B   It is unchanged because the number of particles is unchanged. C   It increases because cooling releases heat to the surroundings. D   It is unchanged because entropy depends only on the container volume.
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A. Removing thermal energy leaves fewer energy allocations accessible to the gas. Fixed volume and fixed particle number do not fix entropy.

Final answer

A

Question B2

spontaneity-signs[1 mark]
For a model process, \Delta H^{\ominus}=-24.6\,\text{kJ mol}^{-1} and \Delta S^{\ominus}=-58.5\,\text{J K}^{-1}\text{ mol}^{-1}. Both remain constant. Which condition gives \Delta G^{\ominus}<0?
A   0<T<147\,\text{K} only B   0<T<421\,\text{K} C   T=421\,\text{K} only D   T>421\,\text{K}
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B. In kJ units, the model gives \Delta G^\circ=−24.6+0.0585T. Its zero is 420.5128 K, reported as 421 K; the unrounded boundary is excluded. The forward standard change is negative below that boundary.

Final answer

B: 0<T<421\,\mathrm K (rounded boundary; the exact zero is excluded).

Question B3

dg-calc[1 mark]
A model reaction has \Delta H^{\ominus}=+12.0\,\text{kJ mol}^{-1} and \Delta S^{\ominus}=+22.0\,\text{J K}^{-1}\text{ mol}^{-1} at 350\,\text{K}. What is \Delta G^{\ominus}?
A   -7.69\times10^{3}\,\text{kJ mol}^{-1} B   -4.3\,\text{kJ mol}^{-1} C   4.3\,\text{kJ mol}^{-1} D   19.7\,\text{kJ mol}^{-1}
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C. Convert 22.0 J K−1 mol−1 to 0.0220 kJ K−1 mol−1. Then 12.0−350(0.0220)=+4.30 kJ mol−1, matching the +4.3 option.

Final answer

C: +4.3\,\mathrm{kJ\,mol^{-1}}.

Question B4

equilibrium-gibbs[1 mark]
A reaction mixture has Q=K/4 at a fixed positive absolute temperature. Both quantities are dimensionless. Which conclusion is justified?
A   The forward reaction is spontaneous, even if \Delta G^{\ominus} is positive. B   The reverse reaction is spontaneous because Q is less than K. C   The mixture is at equilibrium because K is fixed. D   The standard Gibbs energy change must become negative as Q falls.
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A. Eliminate the standard term using \Delta G^\circ=−RT ln K: \Delta G=RT ln(Q/K). Here ln(1/4)<0 and RT>0. A positive standard Gibbs change is compatible with this negative actual change.

Final answer

A

Question B5

entropy-reaction-compare[1 mark]
For these supplied model species, which transformation consumes gas overall and therefore suggests a negative reaction entropy change? A \ce{PQ(g) -> P(g) + Q(g)} B \ce{P(g) + 2Q(g) -> PQ2(l)} C \ce{Q2(l) -> Q2(g)} D \ce{P2Q4(g) -> 2PQ2(g)}
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B. Combining one gaseous P with two gaseous Q amounts makes liquid PQ2: the gaseous amount drops from three to zero. A changes one into two gas amounts, C changes zero into one, and D changes one into two. Read the phase labels before adding the coefficients.

Final answer

B

Question B6

delta-s-calc[2 marks]
For model species, \ce{2X(g) -> X2(g)}. The supplied S^\circ values are 145 for X and 232 for X_2, in \mathrm{J\,K^{-1}\,mol^{-1}}. Calculate \Delta S^\circ.
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The product contributes 232 and the reactants contribute 2(145)=290. Subtract reactants from products: \Delta S^\circ=232-290=-58\,\mathrm{J\,K^{-1}\,mol^{-1}}. The coefficient applies to the whole molar entropy value. An element at ordinary temperature would also contribute its tabulated absolute entropy; the zero-formation convention does not apply to an entropy sum.

Final answer

-58\,\mathrm{J\,K^{-1}\,mol^{-1}}.

Question B7

spontaneity-meaning[2 marks]
A slow process has actual \Delta G=-2.40\,\mathrm{kJ\,mol^{-1}} at constant temperature and pressure. State the sign of the total entropy change and explain why the process can remain slow.
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At constant temperature and pressure, \Delta S_{\rm total}=-\Delta G/T>0 because T>0 and \Delta G<0. The Gibbs sign gives the thermodynamic direction. A large activation energy can still give a small rate.

Final answer

Total entropy increases; an activation barrier can limit the rate.

Question B8

gibbs-graph-data[2 marks]
A straight-line model for \Delta G^\circ against T has gradient -0.0650\,\mathrm{kJ\,K^{-1}\,mol^{-1}} and extrapolated intercept +19.5\,\mathrm{kJ\,mol^{-1}}. State \Delta S^\circ and \Delta H^\circ.
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Compare \Delta G^\circ=\Delta H^\circ-T\Delta S^\circ with a straight line. The gradient is -\Delta S^\circ, so \Delta S^\circ=+0.0650\,\mathrm{kJ\,K^{-1}\,mol^{-1}}=+65.0\,\mathrm{J\,K^{-1}\,mol^{-1}}. The extrapolated intercept is \Delta H^\circ=+19.5\,\mathrm{kJ\,mol^{-1}}; this is a model parameter. The zero-kelvin intercept is an extrapolation; it does not assert that the phase and constant-data assumptions remain valid at zero kelvin.

Final answer

+65.0\,\mathrm{J\,K^{-1}\,mol^{-1}}; +19.5\,\mathrm{kJ\,mol^{-1}}.

Tutorial practice

Question U1

entropy-sign-reason, delta-s-calc[4 marks]
A thermodynamic assessment examines 2\,\mathrm{HCHO(g)}\rightarrow\mathrm{CH_3OH(l)}+\mathrm{CO(g)} at 298.15\,\mathrm{K}. The respective standard molar entropies of methanal, methanol and carbon monoxide are 219, 127 and 198\,\mathrm{J\,K^{-1}\,mol^{-1}}.
(a) Predict the entropy sign from the physical states and gas amounts. Give a reason.   [2]
(b) Obtain \Delta S^{\ominus} for the equation as written.   [2]
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(a) Expect a decrease [1]: the equation consumes two gas amounts and returns one, with the remaining product liquid [1]. (b) Product entropy sum is 127+198=325; reactant sum is 2×219=438 [1]. Their difference is −113 J K−1 mol−1 for the written equation [1].

Final answer

(a) Negative: gas amount decreases. (b) -113\,\mathrm{J\,K^{-1}\,mol^{-1}}.

Question U2

[5 marks]
A model desorption step has \Delta H^{\ominus}=+42.0\,\mathrm{kJ\,mol^{-1}} and \Delta S^{\ominus}=+150\,\mathrm{J\,K^{-1}\,mol^{-1}}. Both are taken as temperature independent.
(a) Find the temperature where \Delta G^{\ominus}=0, in kelvin and degrees Celsius.   [2]
(b) State the temperature inequality for a favourable standard Gibbs change, with a reason.   [2]
(c) Find \Delta G^{\ominus} at 300\,\mathrm{K}.   [1]
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(a) Use 42.0/0.150=280 K [1]; subtract 273.15 to obtain 6.85 °C [1]. This is a model result, rather than a measured transition temperature. (b) T>280 K [1]; the term −0.150T becomes more negative as T rises [1]. (c) 42.0−300(0.150)=−3.00 kJ mol−1 [1].

Final answer

(a) 280\,\mathrm K; 6.85\,{}^\circ\mathrm C. (b) T>280\,\mathrm K; the entropy contribution grows. (c) -3.00\,\mathrm{kJ\,mol^{-1}}.

Question U3

dg-calc, gibbs-chain[4 marks]
For a model liquid rearrangement at 360\,\mathrm{K}, the standard enthalpy change is +12.5\,\mathrm{kJ\,mol^{-1}} and the standard Gibbs change is -3.70\,\mathrm{kJ\,mol^{-1}}.
(a) Obtain \Delta S^{\ominus} in \mathrm{J\,K^{-1}\,mol^{-1}}.   [2]
(b) Explain how an endothermic conversion can have a negative standard Gibbs change at this temperature.   [2]
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(a) Rearrange to \Delta S^\circ=(\Delta H^\circ−\Delta G^\circ)/T: [12.5−(−3.70)]/360=0.0450 kJ K−1 mol−1 [1], or +45.0 J K−1 mol−1 [1]. (b) Here T\Delta S^\circ=+16.2 kJ mol−1 [1]. Subtracting this from +12.5 kJ mol−1 leaves −3.70 kJ mol−1, so the entropy contribution exceeds the enthalpy cost [1].

Final answer

(a) +45.0\,\mathrm{J\,K^{-1}\,mol^{-1}}. (b) T\Delta S^\circ=16.2\,\mathrm{kJ\,mol^{-1}} exceeds the enthalpy cost.

Question U4

equilibrium-gibbs[4 marks]
A molecular isomerisation has \Delta G^{\ominus}=-7.20\,\mathrm{kJ\,mol^{-1}} at 320\,\mathrm{K}. The standard Gibbs change applies to conversion of isomer \mathrm{A} to isomer \mathrm{B}. Take R=8.31\,\mathrm{J\,K^{-1}\,mol^{-1}} and dimensionless equilibrium constants.
(a) Calculate the forward equilibrium constant.   [2]
(b) Determine the reverse constant and the reverse standard Gibbs change at this temperature.   [2]
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(a) ln K=7200/(8.31×320)=2.707581; exponentiation gives K=15.0 [2: logarithm and result]. (b) Use the unrounded K: K_{\rm reverse}=0.0667 [1]. Interchanging initial and final states gives \Delta G^\circ_{\rm reverse}=+7.20 kJ mol−1 [1].

Final answer

(a) 15.0. (b) 0.0667; +7.20\,\mathrm{kJ\,mol^{-1}}.

Question U5

gibbs-graph-data, threshold-temperature[4 marks]
A model gives \Delta G^\circ=+7.80\,\mathrm{kJ\,mol^{-1}} at 260 K and -3.90\,\mathrm{kJ\,mol^{-1}} at 390 K. Treat \Delta H^\circ and \Delta S^\circ as constant. (a) Determine \Delta S^\circ and \Delta H^\circ. [3] (b) Find the temperature interval giving \Delta G^\circ<0 in this model. [1]
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(a) The gradient is (-3.90-7.80)/(390-260)=-0.0900\,\mathrm{kJ\,K^{-1}\,mol^{-1}}. Hence \Delta S^\circ=+90.0\,\mathrm{J\,K^{-1}\,mol^{-1}}. Substitution at 260 K gives \Delta H^\circ=7.80+260(0.0900)=31.2\,\mathrm{kJ\,mol^{-1}}. (b) Set the standard change to zero: T_*=31.2/0.0900=346.666\ldots\,\mathrm K. The line slopes down, so \Delta G^\circ<0 above this boundary, within the constant-data model. The boundary is excluded.

Final answer

(a) +90.0\,\mathrm{J\,K^{-1}\,mol^{-1}}; +31.2\,\mathrm{kJ\,mol^{-1}}. (b) T>347\,\mathrm K (rounded boundary).

Question U6

spontaneity-meaning[4 marks]
A crystallisation releases 8.40\,\mathrm{kJ\,mol^{-1}} into a large bath at 280 K. Its system entropy change is -24.0\,\mathrm{J\,K^{-1}\,mol^{-1}}. (a) Calculate the entropy change of the surroundings. [2] (b) Calculate the total entropy change and the corresponding Gibbs change. [2]
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(a) Heat enters the bath, giving \Delta S_{\rm surroundings}=+8400/280=+30.0\,\mathrm{J\,K^{-1}\,mol^{-1}}. (b) The combined change is -24.0+30.0=+6.00\,\mathrm{J\,K^{-1}\,mol^{-1}}. Thus \Delta G=-280(6.00)=-1680\,\mathrm{J\,mol^{-1}}=-1.68\,\mathrm{kJ\,mol^{-1}}. The surroundings more than offset the system entropy decrease. Use the bath temperature in kelvin because its entropy gained per unit heat depends on absolute temperature.

Final answer

(a) +30.0\,\mathrm{J\,K^{-1}\,mol^{-1}}. (b) +6.00\,\mathrm{J\,K^{-1}\,mol^{-1}}; -1.68\,\mathrm{kJ\,mol^{-1}}.

Question U7

dg-calc[3 marks]
For the supplied balanced model \ce{L + 2M -> N}, formation Gibbs energies are -32.0, -18.0 and -79.0\,\mathrm{kJ\,mol^{-1}} for L, M and N respectively, all in the specified states at one temperature. (a) Calculate the standard reaction Gibbs energy. [2] (b) Give the standard Gibbs change for twice the reverse equation. [1]
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(a) Sum the formation contributions with their coefficients: \Delta G^\circ=-79.0-[-32.0+2(-18.0)]=-11.0\,\mathrm{kJ\,mol^{-1}}. (b) Reversal changes the sign and doubling multiplies the result by two: -2(-11.0)=+22.0\,\mathrm{kJ\,mol^{-1}}. These are changes per written equation. Each formation value refers to one mole of its species, so the coefficient of M must be included in the reactant sum.

Final answer

(a) -11.0\,\mathrm{kJ\,mol^{-1}}. (b) +22.0\,\mathrm{kJ\,mol^{-1}}.

Question U8

temperature-effect, spontaneity-signs[3 marks]
Process P has \Delta H^\circ<0 and \Delta S^\circ>0. Process Q has \Delta H^\circ<0 and \Delta S^\circ<0. Assume their signs and phases persist over the temperature range. (a) Explain why P has \Delta G^\circ<0 throughout the model range. [1] (b) Explain how increasing temperature can reverse the sign of \Delta G^\circ for Q. [2]
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For P, both \Delta H^\circ and -T\Delta S^\circ are negative at every positive temperature in the model. For Q, the enthalpy term is negative but -T\Delta S^\circ is positive and grows with temperature. Q therefore has negative standard Gibbs energy below its positive threshold T_*=\Delta H^\circ/\Delta S^\circ, zero at the threshold and positive above it.

Final answer

P: negative \Delta G^\circ throughout. Q: negative below \Delta H^\circ/\Delta S^\circ, positive above.

Advanced practice

Question A1

entropy-states, entropy-sign-reason[6 marks]
An isomerisation model converts gaseous trans-but-2-ene into gaseous cis-but-2-ene. Both have the molecular formula \mathrm{C_4H_8}. Their standard molar entropies are 297 and 301\,\mathrm{J\,K^{-1}\,mol^{-1}}, respectively.
(a) Determine \Delta S^{\ominus} and explain why equal gas amounts do not force it to be zero.   [3]
(b) The standard enthalpy change is +4\,\mathrm{kJ\,mol^{-1}}. Estimate \Delta G^{\ominus} at 298.15\,\mathrm{K}.   [2]
(c) Explain why a count of gas molecules is insufficient for ranking small entropy differences between molecular isomers.   [1]
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(a) Product minus reactant gives 301−297=+4 J K−1 mol−1 [2: setup and difference]. Molecular energy arrangements differ between these isomers, even with the same gas amount [1]. (b) 4−298.15(0.004)=+2.8074 kJ mol−1 before rounding [1]; report about +3 kJ mol−1 at the coarse precision supplied [1]. (c) Gas amount is a qualitative indicator; this close comparison needs molecular entropy data [1].

Final answer

(a) +4\,\mathrm{J\,K^{-1}\,mol^{-1}}; molecular energy arrangements differ. (b) About +3\,\mathrm{kJ\,mol^{-1}}. (c) Numerical entropy data resolve the small difference.

Question A2

[6 marks]
A model gas capture has \Delta H^{\ominus}=-54.0\,\mathrm{kJ\,mol^{-1}} and \Delta S^{\ominus}=-135\,\mathrm{J\,K^{-1}\,mol^{-1}}. The reverse change regenerates the capture material. Use constant thermodynamic values throughout.
(a) Find the boundary temperature for capture and the temperature condition for a negative capture Gibbs change.   [2]
(b) State the enthalpy and entropy changes for regeneration, with units.   [2]
(c) Obtain the standard Gibbs change for regeneration at 450\,\mathrm{K} and interpret it.   [2]
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(a) −54.0/(−0.135)=400 K [1]; capture has \Delta G^\circ<0 for 0<T<400 K within the model [1]. (b) Reverse both state differences: \Delta H^\circ=+54.0 kJ mol−1 and \Delta S^\circ=+135 J K−1 mol−1 [2]. (c) Regeneration at 450 K has 54.0−450(0.135)=−6.75 kJ mol−1 [1], favouring the reverse route at standard-state composition [1].

Final answer

(a) 400\,\mathrm K; below it. (b) +54.0\,\mathrm{kJ\,mol^{-1}}; +135\,\mathrm{J\,K^{-1}\,mol^{-1}}. (c) -6.75\,\mathrm{kJ\,mol^{-1}}; favourable standard change.

Question A3

gibbs-chain, delta-s-calc[7 marks]
A chloromethane treatment model considers \mathrm{CH_3Cl(g)}+\mathrm{H_2O(g)}\rightarrow\mathrm{CH_3OH(l)}+\mathrm{HCl(g)}. Use the following reference-temperature data.
\begin{array}{c|rr}\text{Species}&S^{\ominus}/\mathrm{J\,K^{-1}\,mol^{-1}}&\Delta H_{\mathrm{f}}^{\ominus}/\mathrm{kJ\,mol^{-1}}\\\hline\mathrm{CH_3Cl(g)}&235&-82\\\mathrm{H_2O(g)}&189&-242\\\mathrm{CH_3OH(l)}&127&-239\\\mathrm{HCl(g)}&187&-92\end{array}
(a) Calculate both \Delta H^{\ominus} and \Delta S^{\ominus}.   [4]
(b) Using constant values of these quantities, estimate \Delta G^{\ominus} at 300\,\mathrm{K}.   [2]
(c) A student infers a favourable standard Gibbs change solely from the exothermic enthalpy. Assess the inference using your results.   [1]
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(a) The formation-enthalpy sums are −331 and −324 kJ mol−1, giving \Delta H^\circ=−7 kJ mol−1 [2]. The entropy sums are 314 and 424 J K−1 mol−1, giving \Delta S^\circ=−110 J K−1 mol−1 [2]. (b) −7−300(−0.110)=+26 kJ mol−1 [2: compatible units/substitution and result]. (c) Exothermicity alone gives the wrong verdict here; the 33 kJ mol−1 entropy penalty outweighs the 7 kJ mol−1 enthalpy benefit [1].

Final answer

(a) -7\,\mathrm{kJ\,mol^{-1}}; -110\,\mathrm{J\,K^{-1}\,mol^{-1}}. (b) +26\,\mathrm{kJ\,mol^{-1}}. (c) Entropy opposes and outweighs the enthalpy benefit.

Question A4

equilibrium-gibbs[7 marks]
A time-resolved simulation follows two dissolved molecular isomers, labelled \mathrm{A} and \mathrm{B}, interconverting at 310\,\mathrm{K}. Their activity ratio is modelled by Q=[\mathrm{B}]/[\mathrm{A}]. Both concentrations have the same units and their ratio is dimensionless. The independent thermodynamic value is K=4.00, and R=8.31\,\mathrm{J\,K^{-1}\,mol^{-1}}.
\begin{array}{c|rrr}\text{Time}/\mathrm{min}&[\mathrm{A}]/\mathrm{mol\,dm^{-3}}&[\mathrm{B}]/\mathrm{mol\,dm^{-3}}&\text{Observation}\\\hline0&0.0800&0.0200&\text{just mixed}\\5&0.0400&0.0600&\text{still changing}\\20&0.0200&0.0800&\text{unchanged thereafter}\end{array}
(a) Find Q at 5\,\mathrm{min} and explain why this ratio cannot yet be called K.   [2]
(b) Calculate \Delta G^{\ominus} at this temperature and the actual reaction \Delta G at 5\,\mathrm{min}.   [3]
(c) Use both the quotient and the final observation to assess whether equilibrium has been reached at the last time.   [2]
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(a) At five minutes Q=0.0600/0.0400=1.50 [1]. This differs from 4.00 and the concentrations are still evolving [1]. (b) −8.31×310×ln(4.00)/1000=−3.57 kJ mol−1 [1]. For that mixture use RT ln(Q/K) [1], yielding −2.53 kJ mol−1 [1]. (c) The last ratio is 0.0800/0.0200=4.00 [1]. Agreement with K, together with stable concentrations, supports equilibrium; actual \Delta G=0 [1].

Final answer

(a) 1.50; composition is still changing. (b) -3.57 and -2.53\,\mathrm{kJ\,mol^{-1}}, respectively. (c) Q=4.00=K and stable concentrations support equilibrium; actual \Delta G=0.

Connected questions

Question X1

gibbs-chain, threshold-temperature[7 marks]
R1.4 + R1.2. A proposed heat-storage step has a Hess route with \Delta H_1^\circ=+64.0 and \Delta H_2^\circ=-16.0\,\mathrm{kJ\,mol^{-1}}; the desired step is route 1 plus route 2. Its \Delta S^\circ is +120\,\mathrm{J\,K^{-1}\,mol^{-1}}, constant over 300–500 K. (a) Use Hess addition to find the step’s \Delta H^\circ. [1] (b) Find \Delta G^\circ at 440 K and the temperature at which its sign changes. [3] (c) At 360 K, evaluate the reverse step. Given w_{\rm out,max}=-\Delta G, find its maximum non-expansion work per mole for a reversible standard-state step. [3]
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(a) Hess addition gives 64.0-16.0=+48.0\,\mathrm{kJ\,mol^{-1}}. (b) Convert entropy to 0.120\,\mathrm{kJ\,K^{-1}\,mol^{-1}}. At 440 K, \Delta G^\circ=48.0-440(0.120)=-4.80\,\mathrm{kJ\,mol^{-1}}; the zero is 48.0/0.120=400\,\mathrm K. (c) At 360 K the forward standard change is 48.0-360(0.120)=+4.80\,\mathrm{kJ\,mol^{-1}}. Reversal gives -4.80\,\mathrm{kJ\,mol^{-1}}. By the supplied work relationship, its reversible maximum non-expansion work output is 4.80\,\mathrm{kJ\,mol^{-1}}.

Final answer

(a) +48.0\,\mathrm{kJ\,mol^{-1}}. (b) -4.80\,\mathrm{kJ\,mol^{-1}} at 440 K; threshold 400 K. (c) Reverse step releases 4.80\,\mathrm{kJ\,mol^{-1}} maximum non-expansion work at 360 K.

Question X2

equilibrium-gibbs[6 marks]
R1.4 + R2.3. An ideal solution contains interconverting isomers A and B at 305 K. For \ce{A <=> B}, K=6.00 and Q=[\mathrm B]/[\mathrm A]. Initially [\mathrm A]=0.0700 and [\mathrm B]=0.0300\,\mathrm{mol\,dm^{-3}}; their sum stays fixed. (a) Calculate Q and state the thermodynamic direction. [1] (b) Calculate the standard and actual reaction Gibbs energies. [3] (c) Find the equilibrium concentration of B. [2]
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(a) Q=0.0300/0.0700=3/7=0.428571\ldots<6.00, so the forward direction is favoured. (b) With R=8.31\,\mathrm{J\,K^{-1}\,mol^{-1}}, \Delta G^\circ=-8.31(305)\ln6/1000=-4.54130\ldots\,\mathrm{kJ\,mol^{-1}}, or -4.54. The actual change is 8.31(305)\ln[(3/7)/6]/1000=-6.68882\ldots\,\mathrm{kJ\,mol^{-1}}, or -6.69. (c) At equilibrium let [\mathrm B]=b; then [\mathrm A]=0.1000-b and b/(0.1000-b)=6. Thus b=0.6000/7=0.085714\ldots\,\mathrm{mol\,dm^{-3}}, or 0.0857.

Final answer

(a) Q=0.429; forward drive. (b) \Delta G^\circ=-4.54; \Delta G=-6.69\,\mathrm{kJ\,mol^{-1}}. (c) [\mathrm B]=0.0857\,\mathrm{mol\,dm^{-3}}.

Question X3

spontaneity-meaning, dg-calc[6 marks]
R1.4 + R2.2. An uncatalysed model reaction has \Delta H^\circ=-32.0\,\mathrm{kJ\,mol^{-1}} and \Delta S^\circ=-60.0\,\mathrm{J\,K^{-1}\,mol^{-1}} at 300 K. A catalyst lowers its forward activation energy by 12.0\,\mathrm{kJ\,mol^{-1}} without changing the Arrhenius prefactor. (a) Calculate \Delta G^\circ. [1] (b) Using the supplied k=A\exp[-E_{\rm a}/(RT)], calculate the rate-constant ratio and state whether K changes. [3] (c) Explain why little product may appear before catalysis and whether catalysis changes the final equilibrium composition. [2]
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(a) \Delta G^\circ=-32.0-300(-0.0600)=-14.0\,\mathrm{kJ\,mol^{-1}}. (b) Divide the supplied Arrhenius expressions; the prefactors cancel, so k_{\rm cat}/k_{\rm uncat}=\exp[12000/(8.31\times300)]=123.159\ldots, or 123. The catalyst alters barriers but leaves the standard Gibbs difference and therefore K unchanged. (c) A small uncatalysed rate can prevent observable conversion despite a favourable standard Gibbs change. The catalyst accelerates approach to the same equilibrium at fixed temperature; it does not shift its composition.

Final answer

(a) -14.0\,\mathrm{kJ\,mol^{-1}}. (b) k_{\rm cat}/k_{\rm uncat}=123; same K. (c) Kinetic barrier; equilibrium composition is unchanged.

From the book

Photon Chemistry: Study Guide for the IB Diploma (SL & HL)

Entropy and spontaneity from the core ideas to full exam questions, with question types, worked examples and graded practice.

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