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Chapter 17: Problem 118

Consider the reaction $$\mathrm{H}_{2}(g)+\mathrm{B} \mathrm{r}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ where \(\Delta H^{\circ}=-103.8 \mathrm{kJ} / \mathrm{mol}\) . In a particular experiment, equal moles of \(\mathrm{H}_{2}(g)\) at 1.00 \(\mathrm{atm}\) and \(\mathrm{Br}_{2}(g)\) at 1.00 atm were mixed in a \(1.00-\mathrm{L}\) flask at \(25^{\circ} \mathrm{C}\) and allowed to reach equilibrium. Then the molecules of \(\mathrm{H}_{2}\) at equilibrium were counted using a very sensitive technique, and \(1.10 \times 10^{13}\) molecules were found. For this reaction, calculate the values of \(K, \Delta G^{\circ},\) and \(\Delta S^{\circ} .\)

Short Answer

Expert verified
For the given reaction, we initially have: Moles of H2 = \((1.10 * 10^{13}) / (6.022 * 10^{23})\) Moles of Br2 = 1 - Moles of H2 Moles of HBr = 2 * Moles of H2 Now calculate the concentrations and the equilibrium constant (K): K = \(\frac{[HBr]^2}{[H2][Br2]}\) Then calculate ΔG° using: ΔG° = -RT * ln(K) Lastly, evaluate ΔS° with the following equation: ΔS° = \(\frac{ΔH° - ΔG°}{T}\) After performing these calculations, you will obtain the values of K, ΔG°, and ΔS° for the given reaction.

Step by step solution

01

Determine the number of moles of each species at equilibrium

Analyze the data from the experiment and find out the number of moles of each species at equilibrium. We know that there are 1.10x10^13 molecules of H2; we can use Avogadro's number to find the number of moles: Moles of H2 = (1.10 * 10^13) / (6.022 x 10^23) Now, we can use the stoichiometry of the reaction to find the balanced equation and calculate the moles of Br2 and HBr at equilibrium as well: 1. Moles of Br2 = Moles_initial - Moles_reaction = Moles_initial - Moles_H2_equilibrium = 1 - Moles of H2 2. Moles of HBr = Moles_initial + Moles_reaction = Moles_initial + 2 * Moles_H2_equilibrium = 2 * Moles of H2 We can also find their concentrations by dividing by the flask's volume (1.00 L), which is the same for all of them.
02

Calculate the equilibrium constant (K)

Using the concentrations of the species at equilibrium, we can calculate the equilibrium constant (K). The balanced equation is: H2(g) + Br2(g) ⇌ 2 HBr(g) So, K = [HBr]^2 / ([H2] * [Br2]) Plug in the concentrations you calculated in step 1 to find K.
03

Calculate ΔG°

Now that we have K, we can use ΔG° = -RT * ln(K) to calculate the standard Gibbs free energy change. The temperature is given in the problem as 25°C, which is 298 K in Kelvin. The gas constant R = 8.314 J/mol K. Plug in these values to get ΔG°.
04

Calculate ΔS°

Finally, we can use the relationship between ΔG°, ΔH°, and ΔS° to determine ΔS°. The standard enthalpy change ΔH° is given in the problem as -103.8 kJ/mol. To find ΔS°, use: ΔG° = ΔH° - T * ΔS° Rearrange the formula for ΔS°: ΔS° = (ΔH° - ΔG°) / T Plug in the values of ΔH°, ΔG°, and T to find ΔS°. Make sure to convert ΔH° to J/mol before dividing. This will give you the values of K, ΔG°, and ΔS° for the reaction.

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Most popular questions from this chapter

Consider the reaction $$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)$$ a. Use \(\Delta G_{f}^{\circ}\) values in Appendix 4 to calculate $\Delta G^{\circ}$ for this reaction. b. Is this reaction spontaneous under standard conditions at 298 $\mathrm{K} ?$ c. The value of \(\Delta H^{\circ}\) for this reaction is 100 . kJ. At what temperatures is this reaction spontaneous at standard conditions? Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from 99.90% to 99.99% purity by the Mond process. The primary reaction involved in the Mond process is $$\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)$$ a. Without referring to Appendix \(4,\) predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperature-dependent. Predict the sign of \(\Delta S_{\text { sum }}\) for this reaction. Explain. c. For $\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{kJ} / \mathrm{mol}\( and \)S^{\circ}=417 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\( at 298 \)\mathrm{K}$ . Using these values and data in Appendix 4 calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. e. The first step of the Mond process involves equilibrating impure nickel with \(\mathrm{CO}(g)\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) at about \(50^{\circ} \mathrm{C} .\) The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the above reaction at \(50 .^{\circ} \mathrm{C}\) f. In the second step of the Mond process, the gaseous \(\mathrm{Ni}(\mathrm{CO})_{4}\) is isolated and heated to $227^{\circ} \mathrm{C}$ . The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the preceding reaction at \(227^{\circ} \mathrm{C}\) . g. Why is temperature increased for the second step of the Mond process? h. The Mond process relies on the volatility of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for its success. Only pressures and temperatures at which \(\mathrm{Ni}(\mathrm{CO})_{4}\) is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a temperature of \(152^{\circ} \mathrm{C}\) . Estimate the maximum pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) that can be attained before the gas will liquefy at \(152^{\circ} \mathrm{C}\) . The boiling point for Nic CO) is \(42^{\circ} \mathrm{C}\) and the enthalpy of vaporization is 29.0 \(\mathrm{kJ} / \mathrm{mol} .\) [Hint: The phase change reaction and the corresponding equilibrium expression are $\mathrm{Ni}(\mathrm{CO})_{4}(l) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) \quad K=P_{\mathrm{NiCO} 4}$ greater than the \(K\) value. \(]\)

Consider the following energy levels, each capable of holding two particles: $$\begin{array}{l}{E=2 \mathrm{kJ}} \\ {E=1 \mathrm{kJ}} \\ {E=0 \quad \underline{X X}}\end{array}$$ Draw all the possible arrangements of the two identical particles (represented by X) in the three energy levels. What total energy is most likely, that is, occurs the greatest number of times? Assume that the particles are indistinguishable from each other.

Monochloroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right)\) can be produced by the direct reaction of ethane gas $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$ with chlorine gas or by the reaction of ethylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) with hydrogen chloride gas. The second reaction gives almost a 100\(\%\) yield of pure $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}$ at a rapid rate without catalysis. The first method requires light as an energy source or the reaction would not occur. Yet \(\Delta G^{\circ}\) for the first reaction is considerably more negative than \(\Delta G^{\circ}\) for the second reaction. Explain how this can be so.

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. $\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$ b. $2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)$ c. \(\mathrm{HCl}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)
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