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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

What is a spontaneous reaction?

Predict the sign of the enthalpy change for the following processes. Give a reason for your prediction.

(a) \({\bf{NaN}}{{\bf{O}}_{\bf{3}}}{\bf{(s)}} \to {\bf{N}}{{\bf{a}}^{\bf{ + }}}{\bf{(aq) + N}}{{\bf{O}}_{\bf{3}}}^{\bf{ - }}{\bf{(aq)}}\)

(b) the freezing of liquid water

(c) \({\bf{C}}{{\bf{O}}_{\bf{2}}}{\bf{(s)}} \to {\bf{C}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

(d) \({\bf{CaCO(s)}} \to {\bf{CaO(s) + C}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

Calculate the standard entropy change for the following process:

\({{\bf{H}}_{\bf{2}}}{\bf{(g) + }}{{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{4}}}{\bf{(g)}} \to {{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{6}}}{\bf{(g)}}\)

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

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

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