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Chapter 19: Problem 28

The following standard Gibbs energy changes are given for \(25^{\circ} \mathrm{C}\) (1) \(\mathrm{SO}_{2}(\mathrm{g})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow \operatorname{COS}(\mathrm{g})+2 \mathrm{CO}_{2}(\mathrm{g})\) \(\Delta G^{\circ}=-246.4 \mathrm{kJ}\) (2) \(\mathrm{CS}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \operatorname{COS}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) \(\Delta G^{\circ}=-41.5 \mathrm{kJ}\) (3) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g}) \longrightarrow \operatorname{COS}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})\) \(\Delta G^{\circ}=+1.4 \mathrm{kJ}\) (4) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})\) \(\Delta G^{\circ}=-28.6 \mathrm{kJ}\) Combine the preceding equations, as necessary, to obtain \(\Delta G^{\circ}\) values for the following reactions. (a) \(\operatorname{COS}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow\) \(\begin{aligned} \mathrm{SO}_{2}(\mathrm{g})+\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) & \Delta G^{\circ}=? \end{aligned}\) (b) \(\cos (g)+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow\) \(\mathrm{SO}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \quad \Delta G^{\circ}=?\) \(\left.+\quad \mathrm{H}_{\mathrm{O}} \mathrm{C}(\mathrm{d})=\mathrm{CO}_{-}^{\circ} \mathrm{G}\right)+\mathrm{H}_{-}^{-} \mathrm{S}(\mathrm{q})\) (c) \(\cos (\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) \(\Delta G^{\circ}=?\) Of reactions (a), (b), and (c), which is spontaneous in the forward direction when reactants and products are present in their standard states?

Short Answer

Expert verified
The standard Gibbs energies for the reactions are \( \Delta G^{\circ}(a) = -125 kJ, \Delta G^{\circ}(b) = -203.5 kJ, and \Delta G^{\circ}(c) = -70.1 kJ\). All three reactions (a), (b), and (c) are spontaneous under standard conditions.

Step by step solution

01

Reaction (a)

Using the principle of conservation of energy and composition, we write a reaction for (a) as a sum of the given reactions. Multiply reaction (1) by 1, reaction (2) by -1, reaction (3) by -1, and reaction (4) by 2. Add these resulting reactions up, and you will get the reaction for (a). The \( \Delta G^{\circ}\) value for the reaction results from the sum of the products of each reaction’s \( \Delta G^{\circ}\) and the respective multipliers. Thus, \( \Delta G^{\circ}(a) = 1* (-246.4 kJ) + -1*(-41.5 kJ) + -1*(1.4 kJ) + 2*(-28.6 kJ) = -125 kJ\)
02

Reaction (b)

Again, we write a reaction for (b) as a sum of the given reactions. Multiply reaction (1) by 1, reaction (2) by -1, and reaction (3) by -1. Summing up, we get the reaction for (b) and its standard Gibbs energy. Thus, \( \Delta G^{\circ}(b) = 1*(-246.4 kJ) + -1*(-41.5 kJ) + -1*(1.4 kJ) = -203.5 kJ\)
03

Reaction (c)

Now, we compose reaction (c) by summing up suitable combinations of the given reactions. Add reaction (2) and reaction (4) to get the reaction (c). As before, calculate \( \Delta G^{\circ}(c) \) as the sum of standard Gibbs energies of reactions (2) and (4), so \( \Delta G^{\circ}(c) = -41.5 kJ + -28.6 kJ = -70.1 kJ\)
04

Determine spontaneity

Finally, we can determine spontaneous reactions. Whenever \( \Delta G^{\circ}\) is negative, the reaction under standard conditions is spontaneous. In this case, all three reactions (a), (b), and (c) have negative \( \Delta G^{\circ}\) values, and thus all three are spontaneous under standard conditions.

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Most popular questions from this chapter

A tabulation of more precise thermodynamic data than are presented in Appendix D lists the following values for \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(298.15 \mathrm{K},\) at a standard state pressure of 1 bar. $$\begin{array}{llll} \hline & \Delta H_{f}^{\circ}, & \Delta G_{f,}^{\circ} & S_{\prime}^{\circ} \\\ & \text { kJ mol }^{-1} & \text {kJ mol }^{-1} & \text {J mol }^{-1} \text {K }^{-1} \\ \hline \mathrm{H}_{2} \mathrm{O}(1) & -285.830 & -237.129 & 69.91 \\ \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) & -241.818 & -228.572 & 188.825 \\ \hline \end{array}$$ (a) Use these data to determine, in two different ways, \(\Delta G^{\circ}\) at \(298.15 \mathrm{K}\) for the vaporization: \(\mathrm{H}_{2} \mathrm{O}(1,1 \mathrm{bar}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g}, 1 \mathrm{bar}) .\) The value you obtain will differ slightly from that on page 838 because here, the standard state pressure is 1 bar, and there, it is 1 atm. (b) Use the result of part (a) to obtain the value of \(K\) for this vaporization and, hence, the vapor pressure of water at \(298.15 \mathrm{K}\) (c) The vapor pressure in part (b) is in the unit bar. Convert the pressure to millimeters of mercury. (d) Start with the value \(\Delta G^{\circ}=8.590 \mathrm{kJ}\), given on page 838 and calculate the vapor pressure of water at 298.15 K in a fashion similar to that in parts (b) and (c). In this way, demonstrate that the results obtained in a thermodynamic calculation do not depend on the convention we choose for the standard state pressure, as long as we use standard state thermodynamic data consistent with that choice.

For each of the following reactions, indicate whether \(\Delta S\) for the reaction should be positive or negative. If it is not possible to determine the sign of \(\Delta S\) from the information given, indicate why. (a) \(\mathrm{CaO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})\) (b) \(2 \mathrm{HgO}(\mathrm{s}) \longrightarrow 2 \mathrm{Hg}(1)+\mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NaCl}(1) \longrightarrow 2 \mathrm{Na}(1)+\mathrm{Cl}_{2}(\mathrm{g})\) (d) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g})\) (e) \(\operatorname{Si}\left(\text { s) }+2 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{SiCl}_{4}(\mathrm{g})\right.\)

Why is \(\Delta G^{\circ}\) such an important property of a chemical reaction, even though the reaction is generally carried out under nonstandard conditions?

For a process to occur spontaneously, (a) the entropy of the system must increase; (b) the entropy of the surroundings must increase; (c) both the entropy of the system and the entropy of the surroundings must increase; (d) the net change in entropy of the system and surroundings considered together must be a positive quantity; (e) the entropy of the universe must remain constant.

Indicate whether each of the following changes represents an increase or a decrease in entropy in a system, and explain your reasoning: (a) the freezing of ethanol; (b) the sublimation of dry ice; (c) the burning of a rocket fuel.
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