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Chapter 16: Q16.4CYL (page 893)

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

Short Answer

Expert verified
  1. Positive; The solid dissolves to give an increase of mobile ions in solution.
  2. Negative; The liquid becomes a more ordered solid.
  3. Positive; The relatively ordered solid becomes a gas.
  4. Positive; There is a net production of one mole of gas.

Step by step solution

01

Define phase transitions

  • For the phase transitions or changes involving the breaking of bonds or intermolecular attractions like hydrogen bonding, heat is added or absorbed and the enthalpy of the system increases and the\(\Delta {\rm{H}}\)is +ve and the reaction is endothermic.
  • The phase changes or transitions involving the formation of bonds or formation of intermolecular interactions, the heat is released and the \(\Delta {\rm{H}}\) is -ve and the reaction is exothermic.
02

Determine the decomposition reaction.

(a) The reaction of converting \({\rm{NaN}}{{\rm{O}}_3}(\;{\rm{s}})\) to its aqueous ions involves the breaking of bond and heat is added up. The reaction is endothermic in nature and the \(\Delta {\rm{H}}\) is +ve.

(b). The freezing of the liquid water is an exothermic reaction as the heat is evolved from the system. The \(\Delta {\rm{H}}\) is -ve.

(c). solid \({\rm{C}}{{\rm{O}}_2}\) (s) sublimes to \({\rm{C}}{{\rm{O}}_2}(\;{\rm{g}})\). The heat is absorbed from the surroundings for the change. The process is endothermic and \(\Delta {\rm{H}}\) is +ve.

(d). solid \({\rm{C}}{{\rm{O}}_2}\) (s) sublimes to \({\rm{C}}{{\rm{O}}_2}(\;{\rm{g}})\). The heat is absorbed from the surroundings for the change. The process is endothermic and \(\Delta {\rm{H}}\) is +ve.

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Most popular questions from this chapter

Although the gas used in the oxyacetylene torch Figure (5.7) is essentially pure acetylene, the heat produced by the combustion of one mole ofacetylene in such a torch is likely, not equal to the enthalpy of combustion of acetylene listed in the table 5.2. Considering the conditions for which the tabulated data are reported. Suggest an explanation.

Determine the standard enthalpy change, entropy change, and free energy change for the conversion of diamond to graphite. Discuss the spontaneity of the conversion with respect to the enthalpy and entropy changes. Explain why diamond spontaneously changing into graphite is not observed.

Consider the decomposition of red mercury(II) oxide under standard state conditions.

\({\bf{2HgO(s, red )}} \to {\bf{2Hg(l) + }}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

(a) Is the decomposition spontaneous under standard state conditions?

(b) Above what temperature does the reaction become spontaneous?

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掳C
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