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Chapter 15: Problem 62

Consider the reaction $$\begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \rightleftharpoons \\ & 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g), \Delta H=-904.4 \mathrm{~kJ} \end{aligned}$$ Does each of the following increase, decrease, or leave unchanged the yield of \(\mathrm{NO}\) at equilibrium? (a) increase \(\left[\mathrm{NH}_{3}\right] ;(\mathbf{b})\) increase \(\left[\mathrm{H}_{2} \mathrm{O}\right] ;(\mathbf{c})\) decrease \(\left[\mathrm{O}_{2}\right] ;(\mathbf{d})\) decrease the volume of the container in which the reaction occurs; (e) add a catalyst; (f) increase temperature.

Short Answer

Expert verified
(a) Increasing the concentration of NH3 will increase the yield of NO. (b) Increasing the concentration of H2O will decrease the yield of NO. (c) Decreasing the concentration of O2 will decrease the yield of NO. (d) Decreasing the volume of the container will decrease the yield of NO. (e) Adding a catalyst will have no effect on the yield of NO. (f) Increasing temperature will decrease the yield of NO.

Step by step solution

01

(a) Increasing the concentration of NH3

Increasing the concentration of NH3 will result in a shift in the equilibrium to the right, favoring the formation of products, including NO. Therefore, the yield of NO will increase.
02

(b) Increasing the concentration of H2O

Increasing the concentration of H2O will result in a shift in the equilibrium to the left, favoring the formation of reactants and consuming some of the products, including NO. Therefore, the yield of NO will decrease.
03

(c) Decreasing the concentration of O2

Decreasing the concentration of O2 will result in a shift in the equilibrium to the left, favoring the formation of reactants and consuming some of the products, including NO. Therefore, the yield of NO will decrease.
04

(d) Decreasing the volume of the container

Decreasing the volume of the container will increase pressure, causing the equilibrium to shift towards the side with fewer moles of gas. In this case, there are 9 moles of gas on the left side of the equation and 10 moles of gas on the right side. Hence, the reaction will shift to the left, favoring the formation of reactants, and the yield of NO will decrease.
05

(e) Adding a catalyst

Adding a catalyst will speed up both the forward and reverse reactions, making the system reach equilibrium more quickly. However, it will have no effect on the position of the equilibrium or the yield of NO.
06

(f) Increasing temperature

Since the reaction is exothermic (ΔH < 0), increasing the temperature will shift the equilibrium to the left, favoring the formation of reactants, as this will counteract the temperature increase and cool the system down. As a result, the yield of NO will decrease.

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Most popular questions from this chapter

For the equilibrium $$2 \operatorname{IBr}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\operatorname{Br}_{2}(g)$$ \(K_{p}=8.5 \times 10^{-3}\) at \(150^{\circ} \mathrm{C}\). If \(5.07 \mathrm{kPa}\) of IBr is placed in a 10.0 -L container, what is the partial pressure of all substances after equilibrium is reached?

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 \(1 \times 10^{-11}\). Assuming that the exhaust gas (total pressure \(101.3 \mathrm{kPa}\) ) 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? Recall that at a fixed pressure and temperature, volume \(\%=\mathrm{mol} \%\).

Consider the reaction $\mathrm{IO}_{4}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\( \)\mathrm{H}_{4} \mathrm{IO}_{6}^{-}(a q) ; K_{c}=3.5 \times 10^{-2} .\( If you start with \)25.0 \mathrm{~mL}\( of a \)0.905 \mathrm{M}\( solution of \)\mathrm{NaIO}_{4}$, and then dilute it with water to \(500.0 \mathrm{~mL},\) what is the concentration of $\mathrm{H}_{4} \mathrm{IO}_{6}^{-}$ at equilibrium?

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