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6th Year OL · Chemistry
Chemical transformations · Chemical thermodynamics

Entropy and spontaneity

Diamond has more energy than graphite, so under standard conditions it "should" turn into graphite by itself. Yet a diamond ring lasts for millennia without changing. Knowing that a reaction is spontaneous tells you nothing about how fast it is. Gibbs free energy and activation energy answer two different questions, and mixing them up is one of the most common mistakes in this topic.

Entropy: the dispersal of energy

Drop a bit of ink into a glass of water and it spreads out evenly. You never see the ink molecules gather themselves back into one corner. What quantity explains why natural processes have a preferred direction?

Particles in a solid, a liquid and a gas, showing entropy increasing as the particles gain freedom of movement solid liquid gas S increases: less order, more arrangements
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For the synthesis reaction of sulfur trioxide: $$2\,\mathrm{SO_2(g)} + \mathrm{O_2(g)} \rightarrow 2\,\mathrm{SO_3(g)}$$ the standard molar entropy values at…

For the synthesis reaction of sulfur trioxide:

$$2\,\mathrm{SO_2(g)} + \mathrm{O_2(g)} \rightarrow 2\,\mathrm{SO_3(g)}$$

the standard molar entropy values at $298\,\text{K}$ are available: $S^\circ[\mathrm{SO_2(g)}] = 248.2\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$, $S^\circ[\mathrm{O_2(g)}] = 205.1\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$ and $S^\circ[\mathrm{SO_3(g)}] = 256.8\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$.

  1. a) Justify qualitatively the expected sign of the standard entropy change of the system, based on the dispersal of matter and the number of available microstates.
  2. b) Calculate the numerical value of $\Delta S^\circ$ for the reaction at $298\,\text{K}$ and confirm the previous prediction.
Solve

Gibbs free energy and spontaneity

A cold pack dissolves ammonium nitrate in water and the pack gets icy cold. The process is endothermic ($\Delta H > 0$), yet it runs by itself and goes to completion. If reactions do not simply "roll downhill" in energy, what quantity balances heat against disorder?

The table below assumes standard conditions and that $\Delta H^\circ$ and $\Delta S^\circ$ stay constant over the temperature range, with no change of state.

The graph shows $\Delta G^\circ = 40 - 0.10\,T$ (in kJ mol⁻¹) for a hypothetical endothermic reaction. Where the line crosses zero, the reaction becomes spontaneous.

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Consider the industrial synthesis reaction of ammonia from its elements at $298\,\text{K}$: $$\mathrm{N_2(g)} + 3\,\mathrm{H_2(g)} \rightarrow 2\,\mathrm{NH_3(g)}$$ For…

Consider the industrial synthesis reaction of ammonia from its elements at $298\,\text{K}$:

$$\mathrm{N_2(g)} + 3\,\mathrm{H_2(g)} \rightarrow 2\,\mathrm{NH_3(g)}$$

For this transformation the following standard thermodynamic data at that temperature are known: reaction enthalpy $\Delta H^\circ = -92.2\,\text{kJ}\times\text{mol}^{-1}$ and entropy change $\Delta S^\circ = -198.7\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$.

  1. a) Calculate the standard Gibbs free energy change ($\Delta G^\circ$) of the reaction at $298\,\text{K}$, paying particular attention to the consistency of the units used.
  2. b) State, justifying your answer using the appropriate thermodynamic criterion, whether the process is spontaneous under standard conditions at that temperature.
Solve
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At $298\,\text{K}$ and $1\,\text{bar}$, the allotropic transformation of diamond carbon into graphite carbon has the following thermodynamic data: $$\mathrm{C(s,…

At $298\,\text{K}$ and $1\,\text{bar}$, the allotropic transformation of diamond carbon into graphite carbon has the following thermodynamic data:

$$\mathrm{C(s, diamante)} \rightarrow \mathrm{C(s, grafito)} \quad \Delta G^\circ_{298} = -2.90\,\text{kJ}\times\text{mol}^{-1}$$

However, diamond jewellery remains intact for millennia without appreciably transforming into graphite.

  1. a) Justify whether the transformation is thermodynamically spontaneous under standard conditions at $298\,\text{K}$.
  2. b) Explain the apparent discrepancy between the negative sign of $\Delta G^\circ$ and the empirical observation, clearly distinguishing between the thermodynamic concept of spontaneity and the rate of the process (kinetic control).
Solve
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For a hypothetical endothermic gas-phase reaction, the standard Gibbs free energy change ($\Delta G^\circ$) has been plotted as a function of the absolute temperature…

For a hypothetical endothermic gas-phase reaction, the standard Gibbs free energy change ($\Delta G^\circ$) has been plotted as a function of the absolute temperature $T$. In the accompanying graph

40 20 0 −20 −40 0 200 600 800 ΔG° / (kJ·mol⁻¹) Temperature T / K ? Gradient = −0.10 kJ/(mol·K)

it can be seen that the line has a y-intercept of $+40\,\text{kJ}\times\text{mol}^{-1}$ and a gradient of $-0.10\,\text{kJ}\times\text{mol}^{-1}\times\text{K}^{-1}$ over the interval analysed ($0$ to $800\,\text{K}$).

  1. a) Using the parameters of the line and the equation $\Delta G^\circ = \Delta H^\circ - T\times\Delta S^\circ$, determine the value of the threshold temperature (marked with "?" at the intersection with the horizontal axis) above which the process is favourable under standard conditions.
  2. b) Discuss the limitations of the usual approximation that treats $\Delta H^\circ$ and $\Delta S^\circ$ as independent of temperature over wide intervals such as the one shown.
Solve
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The thermal decomposition of calcium carbonate is a reaction of great importance in the manufacture of lime: $$\mathrm{CaCO_3(s)} \rightarrow \mathrm{CaO(s)} +…

The thermal decomposition of calcium carbonate is a reaction of great importance in the manufacture of lime:

$$\mathrm{CaCO_3(s)} \rightarrow \mathrm{CaO(s)} + \mathrm{CO_2(g)}$$

At $298\,\text{K}$, the standard values are $\Delta H^\circ = +178.3\,\text{kJ}\times\text{mol}^{-1}$ and $\Delta S^\circ = +160.5\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$.

  1. a) Analyse qualitatively the sign of $\Delta H^\circ$ and $\Delta S^\circ$ using the table of the four possible cases of the Gibbs energy, and deduce how an increase in temperature affects the spontaneity of the process.
  2. b) Calculate the minimum temperature (in degrees Celsius), assuming that $\Delta H^\circ$ and $\Delta S^\circ$ do not vary with temperature, for the decomposition to be spontaneous under standard conditions ($p_{\mathrm{CO_2}} = 1\,\text{bar}$).
Solve

$\Delta G$, $\Delta G^\circ$ and the equilibrium constant

Put only reactants in a flask for a reaction with $\Delta G^\circ > 0$. Some product still forms straight away, even though the reaction looks "unfavourable" on paper. Why does the real driving force change with the mixture, and exactly when does it stop?

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Nitrogen dioxide and its dimer form the following equilibrium in the gas phase at $298.15\,\text{K}$: $$\mathrm{N_2O_4(g)} \rightleftharpoons 2\,\mathrm{NO_2(g)}$$ For…

Nitrogen dioxide and its dimer form the following equilibrium in the gas phase at $298.15\,\text{K}$:

$$\mathrm{N_2O_4(g)} \rightleftharpoons 2\,\mathrm{NO_2(g)}$$

For this system at that temperature, the standard Gibbs free energy change is known to be $\Delta G^\circ = +4.73\,\text{kJ}\times\text{mol}^{-1}$. Data: $R = 8,314\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$.

  1. a) Clearly distinguish the physical meaning of $\Delta G$ and $\Delta G^\circ$. What value does $\Delta G$ take when the system reaches chemical equilibrium?
  2. b) Calculate the value of the standard thermodynamic equilibrium constant $K_p$ at $298.15\,\text{K}$ and explain in which direction a mixture with reaction quotient $Q=1$ would initially evolve. Do not assume that the value of K alone determines all the mole fractions. Use $\Delta G^\circ=-RT\ln K$ and $\Delta G=\Delta G^\circ+RT\ln Q$, with gaseous activities $p_i/p^\circ$ and $p^\circ=1\,\text{bar}$.
Solve
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In iron metallurgy, a classic stage of indirect reduction in the blast furnace is: $$\mathrm{Fe_2O_3(s)} + 3\,\mathrm{CO(g)} \rightarrow 2\,\mathrm{Fe(s)} +…

In iron metallurgy, a classic stage of indirect reduction in the blast furnace is:

$$\mathrm{Fe_2O_3(s)} + 3\,\mathrm{CO(g)} \rightarrow 2\,\mathrm{Fe(s)} + 3\,\mathrm{CO_2(g)}$$

The following standard thermodynamic data at $298\,\text{K}$ are available:

$\Delta H_f^\circ[\mathrm{Fe_2O_3(s)}] = -824.2\,\text{kJ}\times\text{mol}^{-1}$; $\Delta H_f^\circ[\mathrm{CO(g)}] = -110.5\,\text{kJ}\times\text{mol}^{-1}$; $\Delta H_f^\circ[\mathrm{CO_2(g)}] = -393.5\,\text{kJ}\times\text{mol}^{-1}$; $S^\circ[\mathrm{Fe_2O_3(s)}] = 87.4\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$; $S^\circ[\mathrm{CO(g)}] = 197.7\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$; $S^\circ[\mathrm{Fe(s)}] = 27.3\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$; $S^\circ[\mathrm{CO_2(g)}] = 213.7\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$.

  1. a) Calculate $\Delta H^\circ$ and $\Delta S^\circ$ for the reaction at $298\,\text{K}$. Interpret the numerical value obtained for $\Delta S^\circ$ in terms of the number of microstates and the balance of gaseous species.
  2. b) Calculate $\Delta G^\circ$ at $298\,\text{K}$ and at $1000\,\text{K}$, assuming that $\Delta H^\circ$ and $\Delta S^\circ$ remain constant. State, with reasoning, whether the increase in temperature thermodynamically favours or disfavours the standard spontaneity of the reaction.
Solve
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The thermal dissociation of phosphorus pentachloride is studied in the laboratory: $$\mathrm{PCl_5(g)} \rightleftharpoons \mathrm{PCl_3(g)} + \mathrm{Cl_2(g)}$$ The…

The thermal dissociation of phosphorus pentachloride is studied in the laboratory:

$$\mathrm{PCl_5(g)} \rightleftharpoons \mathrm{PCl_3(g)} + \mathrm{Cl_2(g)}$$

The equilibrium constants are determined experimentally at two different temperatures: at $T_1 = 450\,\text{K}$, $K_{p,1} = 1.20$, while at $T_2 = 520\,\text{K}$, $K_{p,2} = 9.70$. Take $R = 8,314\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$ and assume that the standard enthalpy and entropy of the reaction are constant over the interval studied.

  1. a) Calculate the standard enthalpy of reaction ($\Delta H^\circ$) and the standard entropy of reaction ($\Delta S^\circ$) using the integrated form of the van 't Hoff equation provided.
  2. b) In a reactor at $520\,\text{K}$, reactants and products are mixed so that the partial pressures are $p_{\mathrm{PCl_5}} = 2.00\,\text{bar}$, $p_{\mathrm{PCl_3}} = 1.50\,\text{bar}$ and $p_{\mathrm{Cl_2}} = 0.80\,\text{bar}$. Determine the Gibbs free energy change ($\Delta G$) at that specific instant and state whether the standard criterion $\Delta G^\circ$ would serve to predict the direction in which the mixture will evolve. Formulae: $\ln(K_2/K_1)=-(\Delta H^\circ/R)(1/T_2-1/T_1)$, $\Delta G^\circ=-RT\ln K$, $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$ and $\Delta G=\Delta G^\circ+RT\ln Q$. K and Q are dimensionless using $p_i/p^\circ$, with $p^\circ=1\,\text{bar}$.
Solve
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The vaporisation equilibrium of water at a pressure of $1\,\text{atm}$ is described by: $$\mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_2O(g)}$$ The normal boiling point…

The vaporisation equilibrium of water at a pressure of $1\,\text{atm}$ is described by:

$$\mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_2O(g)}$$

The normal boiling point is $373.15\,\text{K}$ ($100\,^\circ\text{C}$), with a standard enthalpy of vaporisation of $\Delta H^\circ_{\text{eb}} = +40.66\,\text{kJ}\times\text{mol}^{-1}$.

  1. a) Calculate the standard entropy change of vaporisation $\Delta S^\circ_{\text{eb}}$ at $373.15\,\text{K}$. Justify its sign in terms of energy dispersal and the increase in molecular microstates.
  2. b) Calculate $\Delta G^\circ$ of vaporisation at $298.15\,\text{K}$, assuming that $\Delta H^\circ$ and $\Delta S^\circ$ do not depend on temperature. In light of this result, explain why liquid water placed in an open dish at $25\,^\circ\text{C}$ can evaporate spontaneously if the surroundings are unsaturated and the air is renewed, even though the value of $\Delta G^\circ$ obtained is positive. In this exercise the gaseous reference state is explicitly set at 1 atm (not 1 bar); the symbol $\circ$ refers to that convention. To discuss the dish, distinguish between the total pressure of the air and the partial pressure of the water vapour.
Solve
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Topic challenge. Is your kitchen efficient? Evaluate the thermodynamics of heating water in your home in the context of the energy transition. 1. Domestic calorimetry:…

Topic challenge. Is your kitchen efficient? Evaluate the thermodynamics of heating water in your home in the context of the energy transition.

  1. Domestic calorimetry: Heat $0.5\text{ kg}$ of water in a saucepan for $3\text{ min}$. Measure $T_i$ and $T_f$ (or take the boiling temperature) and calculate the useful heat absorbed using $Q = m \times c_e \times \Delta T$.

  2. Hess, bonds and the first law: Calculate the molar enthalpy of combustion ($\Delta H^\circ$) of the gas used (butane or methane) by applying Hess's law or bond energies. Estimate the expansion work ($W = -\Delta n_g R T$) of the gases produced and deduce $\Delta U$. Determine the thermal efficiency of your cooker by estimating the energy supplied from the burner's rated power (or the gas consumed) during the $3\text{ min}$.

  3. Entropy and spontaneity: Look up the ambient temperature of your municipality on the Met Éireann website. Using tabulated standard thermodynamic data, calculate $\Delta S^\circ$ and $\Delta G^\circ$ at that temperature and explain the spontaneity of the process.

  4. Technical proposal: Write a reasoned report (one page) with comparative tables, justifying the heat losses and suggesting practical energy-saving measures.

Assessment criteria:

  • Rigour in setting up and calculating thermodynamic variables ($Q$, $W$, $\Delta H^\circ$, $\Delta S^\circ$ and $\Delta G^\circ$).

  • Handling and consistency of the home experimental data and the official meteorological data (Met Éireann).

  • Capacity for critical analysis when interpreting the thermal efficiency and the spontaneity of the system.

Solve

Back to the start: diamond is thermodynamically unstable relative to graphite ($\Delta G^\circ < 0$), but the activation energy is huge, so the change is too slow to see.

Summary

See all the exercises in this section