Consider the system shown in Figure 16.9. What is the change in entropy for the process where all the energy is transferred from the hot object (AB) to the cold object (CD)

Short Answer

Expert verified

The value of \(\Delta {\rm{S}} = 1.91 \times {10^{ - 23}}\;{\rm{J}}/{\rm{K}}\).

Step by step solution

01

Define chemical process

  • Microstate is defined as specific detailed microscopic configuration of a system.
  • Greater the number of microstate available in the system, greater is the entropy of the system. The number of microstates is a measure of the potential disorder of the system.
  • The change in entropy when the microstate changes from particle \(A\) and \(B\) having one unit of energy to particle \(C\) and D having unit of energy each.
  • The change in entropy is given by the expression,
  • \(\begin{array}{c}\Delta S = {S_{\rm{f}}} - {S_{\rm{i}}}\\ = k\ln \left( {\frac{{{W_{\rm{f}}}}}{{{W_{\rm{i}}}}}} \right)\end{array}\)
  • Here, \(\Delta S\) is the change in entropy, \({S_{\rm{f}}}\) and \({S_{\rm{i}}}\) are the final and initial states of the system, \(k\) is the Boltzmann's constant with the value of \(1.38 \times {10^{ - 23}}\;{\rm{J}}/{\rm{K}},{W_{\rm{i}}}\) and \({W_{\rm{f}}}\) are the initial and final microstates of the system.
02

 Determine the decomposition reaction.

The energy is initially associated only with particles \(A\) and \(B\) and final state is with two particles in different boxes.

According to given figure, there are 10 microstates but the energy is initially associated only with particles A and \(B\) is in only one microstate.

Hence,

Final state \({{\rm{W}}_{\rm{f}}} = 4\)

Initial state \({{\rm{W}}_{\rm{i}}} = 1\)

\(\begin{array}{l}{\rm{k}} = 1.38 \times {10^{ - 23}}\;{\rm{J}}/{\rm{K}}\\\Delta {\rm{S}} = {\rm{k}}\ln \left( {\frac{{{{\rm{W}}_{\rm{f}}}}}{{{{\rm{W}}_{\rm{i}}}}}} \right)\\\Delta {\rm{S}} = 1.38 \times {10^{ - 23}}\;{\rm{J}}/{\rm{K}} \times \ln \left( {\frac{4}{1}} \right)\\\Delta {\rm{S}} = 1.91 \times {10^{ - 23}}\;{\rm{J}}/{\rm{K}}\end{array}\)

Therefore, the value of \(\Delta {\rm{S}} = 1.91 \times {10^{ - 23}}\;{\rm{J}}/{\rm{K}}\).

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What happens to \({\bf{\Delta G}}_{{\bf{298}}}^{\bf{^\circ }}\) (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?

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