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Chapter 15: Problem 91

At \(1200 \mathrm{~K}\), the approximate temperature of automobile exhaust gases (Figure 15.15\(), K_{p}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$ is about \(1 \times 10^{-13}\). Assuming that the exhaust gas (total pressure 1 atm \()\) contains \(0.2 \% \mathrm{CO}, 12 \% \mathrm{CO}_{2},\) and \(3 \% \mathrm{O}_{2}\) by volume, is the system at equilibrium with respect to the \(\mathrm{CO}_{2}\) reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst that speeds up the \(\mathrm{CO}_{2}\) reaction?

Short Answer

Expert verified
The reaction is not at equilibrium, as the calculated reaction quotient (Q) is greater than the given equilibrium constant (Kp), \(Q = 8.33 \times 10^{-5}\) and \(K_p = 1 \times 10^{-13}\). Since the reaction proceeds in the reverse direction to reach equilibrium, a catalyst would speed up the conversion of CO and O2 into CO2, resulting in a decrease in the concentration of CO in the automobile exhaust gas.

Step by step solution

01

Write the expression for the Reaction Quotient (Q)

The reaction quotient (Q) for the given reaction can be written as follows: \[Q = \frac{[\mathrm{CO}]^2[\mathrm{O}_{2}]}{[\mathrm{CO}_{2}]^2}\]
02

Calculate molar fractions of gases

We are given the volume percentages of CO, CO2, and O2 in the exhaust gas. We can assume that the molar fraction is equal to the volume fraction because the total pressure is 1 atm and there is no restriction on the volume of each gas. The molar fractions are: \[\mathrm{Molar\,fraction\,of\,CO} = 0.002\] \[\mathrm{Molar\,fraction\,of\,CO_2} = 0.12\] \[\mathrm{Molar\,fraction\,of\,O_2} = 0.03\]
03

Calculate the current reaction quotient (Q) using the given molar fractions

Now we need to substitute the molar fractions of the gases into the reaction quotient expression derived in step 1: \[Q = \frac{(0.002)^2(0.03)}{(0.12)^2} = 8.33 \times 10^{-5}\]
04

Compare Q and Kp

Now, we can compare the calculated value of Q with the given value of Kp: \[Q = 8.33 \times 10^{-5}\] \[K_p = 1 \times 10^{-13}\] Since \(Q > K_p\), the reaction is not at equilibrium. In this case, the reaction will proceed in the reverse direction, converting CO and O2 back into CO2.
05

Analyze the effect of a catalyst on the CO concentration

A catalyst speeds up the reaction without changing the equilibrium constant or the position of the equilibrium. Since we found that the reaction would proceed in the reverse direction to reach equilibrium, a catalyst would speed up the conversion of CO and O2 into CO2, therefore decreasing the concentration of CO in the automobile exhaust gas.

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Most popular questions from this chapter

If \(K_{c}=0.042\) for \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) at \(500 \mathrm{~K}\) what is the value of \(K_{p}\) for this reaction at this temperature?

As shown in Table \(15.2, K_{p}\) for the equilibrium $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ is \(4.51 \times 10^{-5}\) at \(450^{\circ} \mathrm{C}\). For each of the mixtures listed here, indicate whether the mixture is at equilibrium at \(450^{\circ} \mathrm{C}\). If it is not at equilibrium, indicate the direction (toward product or toward reactants) in which the mixture must shift to achieve equilibrium. (a) \(98 \mathrm{~atm} \mathrm{NH}_{3}, 45 \mathrm{~atm} \mathrm{~N}_{2}, 55 \mathrm{~atm} \mathrm{H}_{2}\) (b) \(57 \mathrm{~atm} \mathrm{NH}_{3}, 143 \mathrm{~atm} \mathrm{~N}_{2},\) no \(\mathrm{H}_{2}\) (c) \(13 \mathrm{~atm} \mathrm{NH}_{2} 27 \mathrm{~atm} \mathrm{~N}_{2} 82 \mathrm{~atm} \mathrm{H}_{2}\)

If \(K_{c}=1\) for the equilibrium \(2 \mathrm{~A}(g) \rightleftharpoons \mathrm{B}(g)\), what is the relationship between \([\mathrm{A}]\) and \([\mathrm{B}]\) at equilibrium?

Write the expression for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(3 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g)\) (b) \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\) (c) \(\mathrm{Ni}(\mathrm{CO})_{4}(g) \rightleftharpoons \mathrm{Ni}(s)+4 \mathrm{CO}(g)\) (d) \(\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)\) (e) \(2 \mathrm{Ag}(s)+\mathrm{Zn}^{2+}(a q) \rightleftharpoons 2 \mathrm{Ag}^{+}(a q)+\mathrm{Zn}(s)\) (f) \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)\) (g) \(2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+2 \mathrm{OH}^{-}(a q)\)

Consider the equilibrium \(\mathrm{Na}_{2} \mathrm{O}(s)+\mathrm{SO}_{2}(g) \rightleftharpoons\) \(\mathrm{Na}_{2} \mathrm{SO}_{3}(s)\) (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) All the compounds in this reaction are soluble in water. Rewrite the equilibrium-constant expression in terms of molarities for the aqueous reaction.
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