Question:

How many reactions are non-spontaneous at 300 K. For independent reaction ΔH & ΔS values are given.

Updated On: Aug 2, 2024
  • \(ΔH = –25 kJ/mol, ΔS = –80 J/mol\)

  • \(ΔH = +25 kJ/mol, ΔS = –50 J/mol\)

  • \(ΔH = -22 kJ/mol, ΔS = +50 J/mol\)

  • \(ΔH = –22 kJ/mol, ΔS = 80 J/mol\)

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The Correct Option is B

Solution and Explanation

A reaction is considered spontaneous at a particular temperature when it exhibits a positive change in Gibbs free energy, denoted as \(\Delta G\). The formula for calculating \(\Delta G\) is \(\Delta G = \Delta H - T\Delta S\), where \(\Delta H\) represents the change in enthalpy, and \(\Delta S\) represents the change in entropy. A negative \(\Delta G\) value indicates a spontaneous reaction, while a positive \(\Delta G\) value indicates a non-spontaneous reaction.

\(\mathbf{\triangle H}\)

\(\mathbf{\triangle S}\)

\(\mathbf{\triangle G}\)

\(\mathbf{Spontaneity \,of \,reaction}\)

Negative (exothermic)PositiveNegativeReactions are spontaneous at all temperatures.
Negative (exothermic)NegativeNegative or PositiveReactions become spontaneous at low temperatures.
when \(|T.\triangle S|<|\triangle H|.\)
Positive (endothermic)PositiveNegative or PositiveReactions become spontaneous at low temperatures.
when \(|T.\triangle S|<|\triangle H|.\)
Positive (endothermic)NegativePositiveReactions are non-spontaneous at all temperatures.


Now, let's examine the Gibbs free energy changes for each reaction at a temperature of 300 K:

(A)  \(ΔH = –25 kJ/mol, \,ΔS = –80 J/mol\)

\(\triangle G\,=\triangle H - T\triangle S\)
\(\triangle  G = -25 -300\times (\frac{-80}{1000})\)
         \(= -25 -300 \times (-0.08)\)
         \(= -25 - (-24)\)
         \(= -25 +24\)
\(\triangle  G = -1\)
\(ΔG \text{ is negative, the reaction is spontaneous.}\)

(B) \(ΔH =+25 \,kJ/mol, ΔS = +50 \,J/mol\)

\(\triangle G\,=\triangle H - T\triangle S\)
\(\triangle  G = +25 -300\times (\frac{-50}{1000})\)
         \(= +25 -300\times (-0.05)\)
         \(= +25 -(-15)\)
         \(= +25 +15\)
\(\triangle  G = +40\)
\(ΔG \text{ is positive, the reaction is non-spontaneous.}\) 

(C) \(ΔH = -22\, kJ/mol, ΔS = +50 \,J/mol\)

\(\triangle G\,=\triangle H - T\triangle S\)
\(\triangle  G= -22 -300\times (\frac{50}{1000})\)
         \(=- 22 -300\times (0.05)\)
         \(= -22 -(15)\)
\(\triangle  G= -37\)
\(ΔG \text{ is negative, the reaction is spontaneous.}\)

(D) \(ΔH = –22 \,kJ/mol, ΔS = 80\, J/mol\)

\(\triangle G\,=\triangle H - T\triangle S\)
\(\triangle  G= -22 -300\times (\frac{80}{1000})\)
         \(= -22 -300 \times (0.08)\)
         \(= -22 -24\)
\(\triangle  G = -46\)
\(ΔG \text{ is negative, the reaction is spontaneous.}\)

\(\text{So, The Correct answer is only option (B)}\) \(ΔH = +25 kJ/mol, ΔS = –50 J/mol\).

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Concepts Used:

Laws of Thermodynamics

Thermodynamics in physics is a branch that deals with heat, work and temperature, and their relation to energy, radiation and physical properties of matter.

The First Law of Thermodynamics:

The first law of thermodynamics, also known as the Law of Conservation of Energy, states that energy can neither be created nor destroyed; energy can only be transferred or changed from one form to another. 

The Second Law of Thermodynamics:

The second law of thermodynamics says that the entropy of any isolated system always increases. Isolated systems spontaneously evolve towards thermal equilibrium—the state of maximum entropy of the system. More simply put: the entropy of the universe (the ultimate isolated system) only increases and never decreases.

The Third Law of Thermodynamics:

The third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches absolute zero. The entropy of a system at absolute zero is typically zero, and in all cases is determined only by the number of different ground states it has. Specifically, the entropy of a pure crystalline substance (perfect order) at absolute zero temperature is zero