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Chapter 17: Problem 99

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: $$\mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} \quad \Delta G^{\circ}=-70 \mathrm{kJ}$$ $$\mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} \quad \Delta G^{\circ}=-80 \mathrm{kJ} $$ 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
The equilibrium constant (K) for the reaction HgbO2 + CO ⇌ HgbCO + O2 at 25°C is approximately 56.64.

Step by step solution

01

Calculate the standard free energy change for the desired reaction

We start by calculating the standard free energy change for the reaction of interest: HgbO2 + CO ⇌ HgbCO + O2. We can do this by manipulating the given reactions as follows: 1. Reverse the first reaction: HgbO2 → Hgb + O2; ∆G° = +70 kJ 2. Add the reversed first reaction to the second reaction: (Hgb + O2) + (Hgb + CO) → (HgbO2) + (HgbCO) Now let's calculate the new standard free energy change for the reaction: ∆G°(new) = ∆G°(first reaction, reversed) + ∆G°(second reaction) ∆G°(new) = (+70 kJ) + (-80 kJ) ∆G°(new) = -10 kJ
02

Calculate the equilibrium constant using the standard free energy change

Now that we have the standard free energy change for the desired reaction, we can calculate the equilibrium constant (K) using the following formula: \[ K = e^{(-\Delta G^{\circ} / RT)} \] where R is the gas constant (8.314 J/K·mol) and T is the temperature in Kelvin (25°C = 298K). Plugging the values into the formula, we get: \[ K = e^{(-(-10,000) / (8.314 \times 298))} \]
03

Calculate the equilibrium constant

By simplifying the expression, we obtain the value of the equilibrium constant: \[ K = e^{(10,000 / 2477.612)} \] \[ K = e^{(4.036)} \] \[ K \approx 56.64 \] Therefore, the equilibrium constant (K) for the reaction HgbO2 + CO ⇌ HgbCO + O2 at 25°C is approximately 56.64.

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Most popular questions from this chapter

Liquid water at \(25^{\circ} \mathrm{C}\) is introduced into an evacuated, insulated vessel. Identify the signs of the following thermodynamic functions for the process that occurs: $\Delta H, \Delta S, \Delta T_{\text { water }} \Delta S_{\text { surr }}\( \)\Delta S_{\text { univ }}$

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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.

For the sublimation of iodine at \(25^{\circ} \mathrm{C}\) $$\mathrm{I}_{2}(s) \rightarrow \mathrm{I}_{2}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\) are, respectively, 62 \(\mathrm{kJ}\) and 19 \(\mathrm{kJ}\) . Estimate the temperature at which iodine sublimes. Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

For the reaction at 298 K, $$2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are $-58.03 \mathrm{kJ}\( and \)-176.6 \mathrm{J} / \mathrm{K}$ , respectively. What is the value of \(\Delta G^{\circ}\) at 298 \(\mathrm{K}\) ? Assuming that \(\Delta H^{p}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is \(\Delta G^{\circ}=0 ?\) Is \(\Delta G^{\circ}\) negative above or below this temperature?
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