Question

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

• A. \(ΔH = –25 kJ/mol, ΔS = –80 J/mol\)
• B. \(ΔH = +25 kJ/mol, ΔS = –50 J/mol\)
• C. \(ΔH = -22 kJ/mol, ΔS = +50 J/mol\)
• D. \(ΔH = –22 kJ/mol, ΔS = 80 J/mol\)

Answer 1

The Correct Option is B

To determine the number of non-spontaneous reactions at 300 K given the values of \(\Delta H\) (enthalpy change) and \(\Delta S\) (entropy change) for each reaction, we apply the Gibbs free energy equation:

\(\Delta G = \Delta H - T\Delta S\)

A reaction is non-spontaneous if the Gibbs free energy change (\(\Delta G\)) is positive.

Let's analyze each option:

1. For \(\Delta H = -25 \, \text{kJ/mol}\), \(\Delta S = -80 \, \text{J/mol}\):

   Convert \(\Delta S\) to \(\text{kJ/mol}\):

   \(\Delta S = -80 \, \text{J/mol} = -0.080 \, \text{kJ/mol}\)

   Calculate \(\Delta G\):

   \(\Delta G = -25 - 300 \times (-0.080) = -25 + 24 = -1 \, \text{kJ/mol}\)

   Since \(\Delta G\) is negative, the reaction is spontaneous.

2. For \(\Delta H = +25 \, \text{kJ/mol}\), \(\Delta S = -50 \, \text{J/mol}\):

   Convert \(\Delta S\) to \(\text{kJ/mol}\):

   \(\Delta S = -50 \, \text{J/mol} = -0.050 \, \text{kJ/mol}\)

   Calculate \(\Delta G\):

   \(\Delta G = 25 - 300 \times (-0.050) = 25 + 15 = 40 \, \text{kJ/mol}\)

   Since \(\Delta G\) is positive, the reaction is non-spontaneous.

3. For \(\Delta H = -22 \, \text{kJ/mol}\), \(\Delta S = +50 \, \text{J/mol}\):

   Convert \(\Delta S\) to \(\text{kJ/mol}\):

   \(\Delta S = 50 \, \text{J/mol} = 0.050 \, \text{kJ/mol}\)

   Calculate \(\Delta G\):

   \(\Delta G = -22 - 300 \times (0.050) = -22 - 15 = -37 \, \text{kJ/mol}\)

   Since \(\Delta G\) is negative, the reaction is spontaneous.

4. For \(\Delta H = -22 \, \text{kJ/mol}\), \(\Delta S = 80 \, \text{J/mol}\):

   Convert \(\Delta S\) to \(\text{kJ/mol}\):

   \(\Delta S = 80 \, \text{J/mol} = 0.080 \, \text{kJ/mol}\)

   Calculate \(\Delta G\):

   \(\Delta G = -22 - 300 \times (0.080) = -22 - 24 = -46 \, \text{kJ/mol}\)

   Since \(\Delta G\) is negative, the reaction is spontaneous.

Based on the calculations above, only the reaction with \(\Delta H = +25 \, \text{kJ/mol}\) and \(\Delta S = -50 \, \text{J/mol}\) is non-spontaneous at 300 K. Therefore, there is one non-spontaneous reaction.