Topic 12 · page 8 of 17

The feasibility temperature

At the border of feasibility the total entropy change is zero, which gives T = ΔH ÷ ΔS for the system. Use it to say above or below which temperature a reaction goes.

13 min read12.9(ii)

Read first: Balance and temperature

Where the total is zero 12.9(ii)

On the graph of page 7, ΔStotal\Delta S_\text{total} crosses zero at one temperature. On one side the reaction is feasible, and on the other it is not. To find that temperature, set ΔStotal\Delta S_\text{total} to zero:

ΔSsystem−ΔHT=0⇒T=ΔHΔSsystem \Delta S_\text{system} - \frac{\Delta H}{T} = 0 \qquad\Rightarrow\qquad T = \frac{\Delta H}{\Delta S_\text{system}}

Which side is feasible depends on the signs.

MethodFinding the temperature at which a reaction becomes feasible

  1. Convert ΔH\Delta H to J mol⁻¹, so that it matches ΔSsystem\Delta S_\text{system} in J K⁻¹ mol⁻¹.
  2. Calculate T=ΔH/ΔSsystemT = \Delta H / \Delta S_\text{system} in kelvin. The two minus signs cancel when both are negative.
  3. Convert to °C if asked: °C = K − 273.
  4. Say above (both positive) or below (both negative).
  5. State the assumption: ΔH\Delta H and ΔSsystem\Delta S_\text{system} do not change with temperature.

Worked exampleCalcium carbonate

CaCOX3(s)→CaO(s)+COX2(g)\ce{CaCO3(s) -> CaO(s) + CO2(g)}, ΔH\Delta H = +178 kJ mol⁻¹ and ΔSsystem\Delta S_\text{system} = +160.4 J K⁻¹ mol⁻¹.

  1. T = 178 000 ÷ 160.4 = 1110 K, which is 1110 − 273 = 837 °C.
  2. Both signs are positive, so the reaction is feasible above 1110 K.

Worked exampleAmmonia

NX2(g)+3 HX2(g)→2 NHX3(g)\ce{N2(g) + 3H2(g) -> 2NH3(g)}, ΔH\Delta H = −92 kJ mol⁻¹ and ΔSsystem\Delta S_\text{system} = −198.8 J K⁻¹ mol⁻¹.

  1. T = −92 000 ÷ −198.8 = 463 K (190 °C).
  2. Both signs are negative, so the reaction is feasible below 463 K.
  3. Industry runs this reaction at about 700 K, above 463 K, because lower temperatures are too slow even with a catalyst, a compromise between rate and yield. The calculation says only that products are favoured below 463 K. It says nothing about rate, which is the subject of page 9.
A graph of entropy change against temperature from 300 K to 800 K for ammonia synthesis. The entropy change of the system is a horizontal dashed line at minus 198.8. The entropy change of the surroundings is a curve falling from plus 307 at 300 K towards plus 115 at 800 K. The total crosses zero at 463 K. To the left of that the label says feasible, to the right not feasible.
Ammonia synthesis. Here the surroundings term shrinks as T rises, so the total falls and crosses zero at 463 K. The reaction is feasible below it.

Worked exampleA reaction that is not feasible at room temperature

NX2OX4(g)→2 NOX2(g)\ce{N2O4(g) -> 2NO2(g)}, ΔH\Delta H = +57 kJ mol⁻¹ and ΔSsystem\Delta S_\text{system} = +176 J K⁻¹ mol⁻¹.

  1. At 298 K, ΔStotal\Delta S_\text{total} = 176 − 57 000 ÷ 298 = 176 − 191.3 = −15 J K⁻¹ mol⁻¹, so it is not feasible at 298 K.
  2. T = 57 000 ÷ 176 = 324 K (51 °C). Both signs are positive, so it is feasible above 324 K.

Using ΔG 12.9(ii)

You may also answer with the Gibbs energy change, ΔG\Delta G. The specification allows it but does not require it.

ΔG=ΔH−T ΔSsystem \Delta G = \Delta H - T\,\Delta S_\text{system}

A reaction is feasible when ΔG\Delta G is negative. This is the same test as ΔStotal\Delta S_\text{total} positive, because ΔG=−T ΔStotal\Delta G = -T\,\Delta S_\text{total}. At the border, ΔG=0\Delta G = 0, which gives the same temperature T=ΔH/ΔSsystemT = \Delta H / \Delta S_\text{system}.

Worked exampleGibbs energy for two reactions

Use ΔH\Delta H in kJ and ΔSsystem\Delta S_\text{system} in kJ K⁻¹ mol⁻¹ (divide J by 1000).

  1. Ammonia at 298 K: ΔG\Delta G = −92 − 298 × (−0.1988) = −92 + 59.2 = −32.8 kJ mol⁻¹. Negative, so feasible.
  2. Calcium carbonate at 298 K: ΔG\Delta G = 178 − 298 × 0.1604 = 178 − 47.8 = +130.2 kJ mol⁻¹. Positive, so not feasible. At 1300 K: ΔG\Delta G = 178 − 1300 × 0.1604 = −30.5 kJ mol⁻¹, feasible.

Check yourself

  1. Calculate the temperature, in K and °C, above which NHX4Cl(s)→NHX3(g)+HCl(g)\ce{NH4Cl(s) -> NH3(g) + HCl(g)} is feasible. ΔH\Delta H = +176 kJ mol⁻¹ and the entropies are NH₄Cl(s) 94.6, NH₃(g) 192.3, HCl(g) 186.9 J K⁻¹ mol⁻¹.
  2. A reaction has ΔH\Delta H = −120 kJ mol⁻¹ and ΔSsystem\Delta S_\text{system} = −250 J K⁻¹ mol⁻¹ (illustrative values). Find the temperature at which ΔStotal\Delta S_\text{total} = 0, and state whether the reaction is feasible above or below it.
  3. For NX2OX4(g)→2 NOX2(g)\ce{N2O4(g) -> 2NO2(g)} calculate ΔG\Delta G at 298 K, using ΔH\Delta H = +57 kJ mol⁻¹ and ΔSsystem\Delta S_\text{system} = +176 J K⁻¹ mol⁻¹. Is it feasible?
  4. A student calculates the feasibility temperature for calcium carbonate as 1.11 K. Find the error.
  5. State the assumption made when you calculate a feasibility temperature this way.
  6. Explain, using ΔSsurroundings\Delta S_\text{surroundings}, why a reaction with both ΔH\Delta H and ΔSsystem\Delta S_\text{system} positive becomes feasible at high temperature.

Answers cover these until you have tried the questions

  1. ΔSsystem\Delta S_\text{system} = (192.3 + 186.9) − 94.6 = +284.6 J K⁻¹ mol⁻¹. T = 176 000 ÷ 284.6 = 618 K, which is 618 − 273 = 345 °C. Both signs are positive, so feasible above this temperature.
  2. T = −120 000 ÷ −250 = 480 K (207 °C). Both signs are negative, so the reaction is feasible below 480 K.
  3. ΔG\Delta G = 57 − 298 × 0.176 = 57 − 52.4 = +4.55 kJ mol⁻¹. Positive, so not feasible at 298 K, which agrees with ΔStotal\Delta S_\text{total} = −15 J K⁻¹ mol⁻¹ found earlier.
  4. The student divided 178 (kJ) by 160.4 (J K⁻¹ mol⁻¹) without converting. ΔH\Delta H must be 178 000 J mol⁻¹, giving 178 000 ÷ 160.4 = 1110 K.
  5. ΔH\Delta H and ΔSsystem\Delta S_\text{system} are assumed not to change with temperature.
  6. ΔSsurroundings\Delta S_\text{surroundings} = −ΔH ÷ T is negative because ΔH is positive. As T increases, the size of this negative term decreases, so the positive ΔSsystem\Delta S_\text{system} eventually outweighs it and ΔStotal\Delta S_\text{total} becomes positive.