Question at position 10 In 1774 Joseph Priestly prepared the element oxygen by heating mercury(II) oxide 2HgO(s) → 2Hg(l) + O2 (g) For this reaction ΔH° = 90.84 kJ and ΔS° = 108 J/K. In what temperature range is this reaction spontaneous? temperatures above 568 °C the reaction is not spontaneous in any temperature range the reaction is spontaneous at all temperaturestemperatures below 568 °CSingle choice
A
temperatures above 568 °C
B
the reaction is not spontaneous in any temperature range
C
the reaction is spontaneous at all temperatures
D
temperatures below 568 °C

Question

Question at position 10 In 1774 Joseph Priestly prepared the element oxygen by heating mercury(II) oxide 2HgO(s) → 2Hg(l) + O2 (g) For this reaction ΔH° = 90.84 kJ and ΔS° = 108 J/K. In what temperature range is this reaction spontaneous? temperatures above 568 °C the reaction is not spontaneous in any temperature range the reaction is spontaneous at all temperaturestemperatures below 568 °CSingle choice

A

temperatures above 568 °C

B

the reaction is not spontaneous in any temperature range

C

the reaction is spontaneous at all temperatures

D

temperatures below 568 °C

BlackTom AI · Accepted Answer

Gibbs free energy ΔG = ΔH − TΔS, so spontaneity depends on the signs of ΔH and ΔS and the temperature. 1) Relevant knowledge points and concepts - ΔG determines spontaneity: ΔG < 0 → spontaneous, ΔG = 0 → equilibrium, ΔG > 0 → nonspontaneous. - For a process with given standard enthalpy ΔH° and entropy ΔS°, ΔG°(T) = ΔH° − TΔS°. - Units must match (J for ΔH when ΔS is J/K), and temperature in kelvin. 2) Complete reasoning process step by step - Given: ΔH° = +90.84 kJ and ΔS° = +108 J/K. Both signs are positive. - Convert ΔH° to joules: ΔH° = 90.84 kJ × 1000 J/kJ = 90,840 J. - Set ΔG° = 0 to find the threshold temperature where spontaneity changes: 0 = ΔH° − T·ΔS° → T = ΔH° / ΔS°. 3) Calculations with intermediate steps - T = 90,840 J / 108 J·K−1 = 90,840 / 108 = 841.111... K. - Convert to degrees Celsius: T(°C) = 841.111 K − 273.15 = 567.961 °C ≈ 568 °C. 4) Why the correct answer is correct - Since both ΔH° and ΔS° are positive, ΔG° = ΔH° − TΔS° will be positive at low T (nonspontaneous) and become negative at sufficiently high T. The changeover occurs at T ≈ 841.11 K (≈ 568 °C). For T > 841.11 K the term TΔS° exceeds ΔH°, making ΔG° < 0 and the reaction spontaneous. Thus the reaction is spontaneous at temperatures above ≈568 °C. 5) Why the other options are incorrect - "the reaction is not spontaneous in any temperature range": incorrect because at high T the −TΔS term can outweigh ΔH, making ΔG negative. - "the reaction is spontaneous at all temperatures": incorrect because at low T ΔG = ΔH − TΔS is positive (ΔH positive, small TΔS), so nonspontaneous at low T. - "temperatures below 568 °C": incorrect because below the calculated threshold ΔG > 0, so the reaction is nonspontaneous there. 6) Conclusion confirming final answer - Therefore the reaction is spontaneous at temperatures above 568 °C, which matches the provided correct answer.

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