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Chapter 15: Problem 62

Consider \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g),\) \(\Delta H=-904.4 \mathrm{~kJ} .\) How does each of the following changes affect the yield of \(\mathrm{NO}\) at equilibrium? Answer increase, decrease, or no change: (a) increase \(\left[\mathrm{NH}_{3}\right] ;(\mathbf{b})\) increase \(\left[\mathrm{H}_{2} \mathrm{O}\right] ;(\mathrm{c})\) decrease \(\left[\mathrm{O}_{2}\right]\) (d) decrease the volume of the container in which the reaction occurs; (e) add a catalyst; (f) increase temperature.

Short Answer

Expert verified
(a) Increase the concentration of NH3: The yield of NO at equilibrium will increase. (b) Increase the concentration of H2O: The yield of NO at equilibrium will decrease. (c) Decrease the concentration of O2: The yield of NO at equilibrium will decrease. (d) Decrease the volume of the container: The yield of NO at equilibrium will decrease. (e) Add a catalyst: There will be no change in the yield of NO at equilibrium. (f) Increase the temperature: The yield of NO at equilibrium will decrease.

Step by step solution

01

a) Increase [NH3]

By increasing the concentration of NH3, we are disturbing the equilibrium. According to Le Chatelier's principle, the reaction will counteract this change by consuming the excess NH3 and producing more NO. Hence, increasing [NH3] will increase the yield of NO at equilibrium.
02

b) Increase [H2O]

By increasing the concentration of H2O, the equilibrium will again be disturbed. This time, Le Chatelier's principle will cause the reaction to shift in the opposite direction, consuming more H2O and producing more reactants (NH3 and O2). Therefore, increasing [H2O] will decrease the yield of NO at equilibrium.
03

c) Decrease [O2]

Decreasing the concentration of O2 will disturb the equilibrium, and according to Le Chatelier's principle, the reaction will shift in the opposite direction to generate more O2. This means the reaction will move towards the reactants side, consuming NO and producing more NH3 and O2. Hence, decreasing [O2] will decrease the yield of NO at equilibrium.
04

d) Decrease the volume of the container

A decrease in volume will increase the pressure of the system. Le Chatelier's principle dictates that the system will react to minimize the pressure increase by shifting the equilibrium towards the side with fewer gas molecules. In this case, there are 9 gas molecules on the reactant side and 10 on the product side. Therefore, decreasing the volume of the container will shift the reaction towards the reactants, decreasing the yield of NO at equilibrium.
05

e) Add a catalyst

Adding a catalyst to a reaction at equilibrium has no effect on the position of the equilibrium, as it simply speeds up the rate at which the reaction reaches equilibrium. Thus, adding a catalyst will not change the yield of NO at equilibrium.
06

f) Increase the temperature

Since the reaction is exothermic (\(\Delta H = -904.4~kJ\)), increasing the temperature will favor the endothermic (reverse) reaction according to Le Chatelier's principle, which means that more reactants (NH3 and O2) will be produced, and the yield of NO at equilibrium will decrease.

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Most popular questions from this chapter

At \(900 \mathrm{~K}\) the following reaction has \(K_{p}=0.345\) : $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ In an equilibrium mixture the partial pressures of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are 0.135 atm and 0.455 atm, respectively. What is the equilibrium partial pressure of \(\mathrm{SO}_{3}\) in the mixture?

When \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a \(2.00-\mathrm{L}\) flask at \(303 \mathrm{~K}\), \(56 \%\) of the \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) : $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ (a) Calculate \(K_{c}\) for this reaction at this temperature. (b) Calculate \(K_{p}\) for this reaction at \(303 \mathrm{~K}\). (c) Repeat these calculations for \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in a \(15.00-\mathrm{L}\) vessel at \(303 \mathrm{~K}\).

A mixture of \(\mathrm{H}_{2}, \mathrm{~S},\) and \(\mathrm{H}_{2} \mathrm{~S}\) is held in a \(1.0-\mathrm{L}\) vessel at \(90{ }^{\circ} \mathrm{C}\) and reacts according to the equation: $$\mathrm{H}_{2}(g)+\mathrm{S}(s) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)$$ At equilibrium the mixture contains \(0.46 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) and \(0.40 \mathrm{~g}\) \(\mathrm{H}_{2}\) (a) Write the equilibrium-constant expression for this reaction. (b) What is the value of \(K_{c}\) for the reaction at this temperature? (c) Why can we ignore the amount of \(\mathrm{S}\) when doing the calculation in part (b)?

For the equilibrium $$ \mathrm{PH}_{3} \mathrm{BCl}_{3}(s) \rightleftharpoons \mathrm{PH}_{3}(g)+\mathrm{BCl}_{3}(g) $$ \(K_{p}=0.052\) at \(60^{\circ} \mathrm{C}\). (a) Calculate \(K_{c}\) (b) After \(3.00 \mathrm{~g}\) of solid \(\mathrm{PH}_{3} \mathrm{BCl}_{3}\) is added to a closed \(1.500\) - \(\mathrm{L}\) vessel at \(60{ }^{\circ} \mathrm{C}\), the vessel is charged with \(0.0500 \mathrm{~g}\) of \(\mathrm{BCl}_{3}(\mathrm{~g})\). What is the equilibrium concentration of \(\mathrm{PH}_{3}\) ?

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen: \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) .\) An equilibrium mixture in a \(2.00-\mathrm{L}\) vessel is found to contain \(0.0406 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{OH},\) \(0.170 \mathrm{~mol} \mathrm{CO},\) and \(0.302 \mathrm{~mol} \mathrm{H}_{2}\) at \(500 \mathrm{~K} .\) Calculate \(K_{c}\) at this temperature.
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