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Chapter 12: Problem 61

Hemoglobin (Hb) binds to both oxygen and carbon monoxide. When the carbon monoxide replaces the oxygen in an organism, the following reaction occurs: $$\mathrm{HbO}_{2}(a q)+\mathrm{CO}(g) \rightleftharpoons \mathrm{HbCO}(a q)+\mathrm{O}_{2}(g)$$ At \(37^{\circ} \mathrm{C}, K\) is about 200 . When equal concentrations of \(\mathrm{HbO}_{2}\) and \(\mathrm{HbCO}\) are present, the effect of CO inhalation is fatal. Assuming \(\mathrm{P}_{\mathrm{O}_{2}}=0.21 \mathrm{~atm}\), what is \(\mathrm{P}_{\mathrm{CO}}\) when \(\left[\mathrm{HbO}_{2}\right]=[\mathrm{HbCO}] ?\)

Short Answer

Expert verified
Answer: The partial pressure of carbon monoxide when the concentrations of HbO2 and HbCO are equal is 0.00105 atm.

Step by step solution

01

Write the expression for the equilibrium constant (K)

Based on the given reaction, we can write the equilibrium constant K as follows: $$K=\frac{[\mathrm{HbCO}][\mathrm{O}_{2}]}{[\mathrm{HbO}_{2}][\mathrm{CO}]}$$
02

Use the given information to set up the equation

We are given that the concentrations of HbO2 and HbCO are equal, so we can rewrite the K expression as: $$K=\frac{[\mathrm{HbCO}][\mathrm{P}_{\mathrm{O}_{2}}]}{[\mathrm{HbO}_{2}][\mathrm{P}_{\mathrm{CO}}]}$$ Since the concentrations of HbO2 and HbCO are equal, we can replace both with a variable x. $$K=\frac{x[\mathrm{P}_{\mathrm{O}_{2}}]}{x\mathrm{[\mathrm{P}_{\mathrm{CO}}]}}$$
03

Cancel out x

Now, since both the numerator and denominator have an x term, we can cancel them out: $$K=\frac{\mathrm{P}_{\mathrm{O}_{2}}}{\mathrm{P}_{\mathrm{CO}}}$$
04

Solve for P_CO

To find P_CO, we can rearrange the equation to isolate the term: $$\mathrm{P}_{\mathrm{CO}}=\frac{\mathrm{P}_{\mathrm{O}_{2}}}{K}$$
05

Substitute the given values

Now, we can substitute the given values for K and P_O2 into the equation: $$\mathrm{P}_{\mathrm{CO}}=\frac{0.21\ \mathrm{atm}}{200}$$
06

Calculate P_CO

Finally, we can calculate the value of P_CO: $$\mathrm{P}_{\mathrm{CO}}= 0.00105\ \mathrm{atm}$$ Therefore, the partial pressure of carbon monoxide when the concentrations of HbO2 and HbCO are equal is 0.00105 atm.

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Most popular questions from this chapter

Benzaldehyde, a flavoring agent, is obtained by the dehydrogenation of benzyl alcohol. $$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}(g)+\mathrm{H}_{2}(g)$$ \(K\) for the reaction at \(250^{\circ} \mathrm{C}\) is \(0.56\). If \(1.50 \mathrm{~g}\) of benzyl alcohol is placed in a 2.0-L flask and heated to \(250^{\circ} \mathrm{C}\), (a) what is the partial pressure of the benzaldehyde when equilibrium is established? (b) how many grams of benzyl alcohol remain at equilibrium?

For the system $$\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)$$ at \(1000 \mathrm{~K}, K=0.45 .\) Sulfur trioxide, originally at \(1.00 \mathrm{~atm}\) pressure, partially dissociates to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) at \(1000 \mathrm{~K}\). What is its partial pressure at equilibrium?

A sealed flask has \(0.541\) atm of \(\mathrm{SO}_{3}\) at \(1000 \mathrm{~K}\). The following equilibrium is established. $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ At equilibrium, the partial pressure of oxygen is measured to be \(0.216\) atm. Calculate \(K\) for the decomposition of \(\mathrm{SO}_{3}\) at \(1000 \mathrm{~K}\).

Given the following reactions and their equilibrium constants, $$\begin{array}{cl}\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) & K=2.4 \times 10^{-9} \\ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) & K=8.8 \times 10^{-13} \end{array}$$ calculate \(K\) for the reaction $$\mathrm{C}(s)+\mathrm{CO}_{2}(g)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{COCl}_{2}(g)$$

The following data are for the system $$\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)$$ $$\begin{array}{lcccccc}\hline \text { Time (s) } & 0 & 20 & 40 & 60 & 80 & 100 \\ P_{\mathrm{A}} \text { (atm) } & 1.00 & 0.83 & 0.72 & 0.65 & 0.62 & 0.62 \\ P_{B} \text { (atm) } & 0.00 & 0.34 & 0.56 & 0.70 & 0.76 & 0.76 \\\\\hline\end{array}$$ (a) How long does it take the system to reach equilibrium? (b) How does the rate of the forward reaction compare with the rate of the reverse reaction after 30 s? After 90 s?
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