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Chapter 17: Problem 89

Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does \(\mathrm{O}_{2}\). Consider the following reactions and approximate standard free energy changes: $$ \begin{array}{cl} \mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\ \mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{array} $$ Using these data, estimate the equilibrium constant value at \(25^{\circ} \mathrm{C}\) for the following reaction: $$ \mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2} $$

Short Answer

Expert verified
For the reaction HgbO₂ + CO ⇌ HgbCO + O₂ at 25°C, the equilibrium constant (K) value is estimated to be approximately 55.6.

Step by step solution

01

Identify the given information

: We are given the following standard free energy changes for two separate reactions: 1. Hgb + O₂ → HgbO₂ with ∆G° = -70 kJ 2. Hgb + CO → HgbCO with ∆G° = -80 kJ And we are asked to determine the equilibrium constant value for the third reaction: 3. HgbO₂ + CO ⇌ HgbCO + O₂
02

Find the required reaction by combining given reactions

: To find the desired reaction (the third reaction), we need to reverse the first reaction and add it to the second reaction: - Reverse Reaction 1: HgbO₂ → Hgb + O₂ with ∆G° = 70 kJ - Reaction 2: Hgb + CO → HgbCO with ∆G° = -80 kJ - Combine: HgbO₂ + CO ⇌ HgbCO + O₂ with ∆G° = (-80+70) kJ = -10 kJ Now that we have found the ΔG° for the required reaction, we can find the equilibrium constant value.
03

Determine the equilibrium constant using Gibbs free energy formula

: We can use the following equation to relate the standard free energy change (ΔG°) and the equilibrium constant (K) at a given temperature: ΔG° = -RT ln(K) Where: - ΔG° is the standard free energy change - R is the gas constant (8.314 J/mol*K) - T is the temperature in Kelvin - K is the equilibrium constant First, we need to convert the given temperature from Celsius to Kelvin: T = 25°C + 273.15 = 298.15 K Now, let's calculate the equilibrium constant by rearranging the formula and plugging in the values: ln(K) = -ΔG° / (RT) ln(K) = -(-10 × 10³) / (8.314 × 298.15) ln(K) ≈ 4.02 To find the K value, we need to take the exponential of both sides: K ≈ e⁴.⁰² ≈ 55.6
04

State the equilibrium constant value for the reaction

: For the reaction HgbO₂ + CO ⇌ HgbCO + O₂ at 25°C, the equilibrium constant (K) value is estimated to be approximately 55.6.

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Most popular questions from this chapter

You have a \(1.00\) - \(L\) sample of hot water \(\left(90.0^{\circ} \mathrm{C}\right)\) sitting open in a \(25.0^{\circ} \mathrm{C}\) room. Eventually the water cools to \(25.0^{\circ} \mathrm{C}\) while the temperature of the room remains unchanged. Calculate \(\Delta S_{\text {surr }}\) for this process. Assume the density of water is \(1.00 \mathrm{~g} / \mathrm{cm}^{3}\) over this temperature range, and the heat capacity of water is constant over this temperature range and equal to \(75.4 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\).

For the reaction $$ \mathrm{SF}_{4}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{SF}_{6}(g) $$ the value of \(\Delta G^{\circ}\) is \(-374 \mathrm{~kJ}\). Use this value and data from Appendix 4 to calculate the value of \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{SF}_{4}(g)\).

The equilibrium constant \(K\) for the reaction $$ 2 \mathrm{Cl}(g) \rightleftharpoons \mathrm{Cl}_{2}(g) $$ was measured as a function of temperature (Kelvin). A graph of \(\ln (K)\) versus \(1 / T\) for this reaction gives a straight line with a slope of \(1.352 \times 10^{4} \mathrm{~K}\) and a \(y\) -intercept of \(-14.51\). Determine the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. See Exercise 79

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 temperaturedependent. Predict the sign of \(\Delta S_{\text {surr }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{~kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{~J} / \mathrm{K}\). 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 \(\mathrm{Ni}(\mathrm{CO})_{4}\) 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 $$ \mathrm{Ni}(\mathrm{CO})_{4}(l) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) \quad K=P_{\mathrm{Ni}(\mathrm{CO})} $$ \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) will liquefy when the pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is greater than the \(K\) value.]

The following reaction occurs in pure water: $$ \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ which is often abbreviated as $$ \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ For this reaction, \(\Delta G^{\circ}=79.9 \mathrm{~kJ} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C}\). Calculate the value of \(\Delta G\) for this reaction at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{OH}^{-}\right]=0.15 \mathrm{M}\) and \(\left[\mathrm{H}^{+}\right]=0.71 M\).
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