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Chapter 16: Q16.2CYL (page 888)

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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Most popular questions from this chapter

An important source of copper is from the copper ore, chalcocite, a form of copper(I) sulfide. When heated, the \({\bf{C}}{{\bf{u}}_{\bf{2}}}{\bf{S}}\) decomposes to form copper and sulfur described by the following equation:

\({\bf{C}}{{\bf{u}}_{\bf{2}}}{\bf{\;S(s)}} \to {\bf{Cu(s) + S(s)}}\)

(a) Determine \({\bf{\Delta G}}_{{\bf{298}}}^{\bf{^\circ }}\)for the decomposition of \({\bf{C}}{{\bf{u}}_{\bf{2}}}{\bf{S(\;s)}}\).

(b) The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine\({\bf{\Delta G}}_{{\bf{298}}}^{\bf{^\circ }}\)for the process.

(c) The production of copper from chalcocite is performed by roasting the \({\bf{C}}{{\bf{u}}_{\bf{2}}}{\bf{S}}\) in air to produce the \({\bf{Cu}}\). By combining the equations from Parts (a) and (b), write the equation that describes the roasting of the chalcocite, and explain why coupling these reactions together makes for a more efficient process for the production of the copper.

Without doing a numerical calculation, determine which of the following will reduce the free energy change for the reaction, that is, make it less positive or more negative, when the temperature is increased. Explain.

(a) \({{\bf{N}}_{\bf{2}}}{\bf{(g) + 3}}{{\bf{H}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2N}}{{\bf{H}}_{\bf{3}}}{\bf{(g)}}\)

(b) \({\bf{HCl(g) + N}}{{\bf{H}}_{\bf{3}}}{\bf{(g)}} \to {\bf{N}}{{\bf{H}}_{\bf{4}}}{\bf{Cl(s)}}\)

(c) \({\left( {{\bf{N}}{{\bf{H}}_{\bf{4}}}} \right)_{\bf{2}}}{\bf{C}}{{\bf{r}}_{\bf{2}}}{{\bf{O}}_{\bf{7}}}{\bf{(s)}} \to {\bf{C}}{{\bf{r}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(s) + 4}}{{\bf{H}}_{\bf{2}}}{\bf{O(g) + }}{{\bf{N}}_{\bf{2}}}{\bf{(g)}}\)

(d) \({\bf{2Fe(s) + 3}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{F}}{{\bf{e}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(s)}}\)

One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):

\({\text{G}}6{\text{P}} \rightleftharpoons {\text{F}}6{\text{P}}\;\;\;\Delta G_{298}^\circ = 1.7\;{\text{kJ}}\)

(a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?

(b) Standard thermodynamic conditions imply the concentrations of G6P and F6P to be \(1M\), however, in a typical cell, they are not even close to these values. Calculate \(\Delta G\)when the concentrations of G6P and F6P are \(120\mu M\) and \(28\mu M\)respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37°C

Under what conditions is \({{\bf{N}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(g)}} \to {\bf{NO(g) + N}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\) spontaneous?

Popular chemical hand warmers generate heat by the air-oxidation of iron:\({\bf{4Fe(s) + 3}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2F}}{{\bf{e}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(s)}}\).How does the spontaneity of this process depend upon temperature?
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