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Chapter 12: Problem 84

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?

Short Answer

Expert verified
a) The partial pressure of benzaldehyde at equilibrium is approximately 0.74 atm. b) The mass of benzyl alcohol remaining at equilibrium is approximately 0.605 grams.

Step by step solution

01

Calculate the initial moles of benzyl alcohol

Convert the given mass of benzyl alcohol to moles using its molar mass (108.14 g/mol). $$\mathrm{Moles~of~benzyl~alcohol} = \frac{1.50~\mathrm{g}}{108.14~\mathrm{g/mol}} = 0.0139~\mathrm{mol}$$ Initially, no benzaldehyde or hydrogen are present, so their initial moles will be equal to 0.
02

Set up the ICE table

Using the initial moles calculated in step 1, create an ICE table to track the change in moles during the reaction. We will use 'x' to represent the change in moles. | Species | Initial moles | Change in moles | Equilibrium moles | |------------------|---------------|-----------------|-------------------| | Benzyl alcohol | 0.0139 | -x | 0.0139 - x | | Benzaldehyde | 0 | x | x | | Hydrogen | 0 | x | x |
03

Express the equilibrium constant in terms of moles and volume

Using the given value of K and the equilibrium expressions from the ICE table, we can write the equilibrium equation as follows: $$K = \frac{[\mathrm{C_{6}H_{5}CHO}][\mathrm{H_2}]}{[\mathrm{C_{6}H_{5}CH_2OH}]}$$ Convert the moles of each species to concentration by dividing them by volume (2 L): $$K = \frac{\frac{x}{2} \cdot \frac{x}{2}}{\frac{0.0139 - x}{2}}$$
04

Solve for the change in moles 'x'

Substitute the given value of K (0.56) into the equilibrium equation and solve for the change in moles 'x': $$0.56 = \frac{\frac{x^2}{4}}{\frac{0.0139 - x}{2}}$$ Solve the equation for x: $$x \approx 0.00832$$
05

Calculate the partial pressure of benzaldehyde

Convert the equilibrium moles of benzaldehyde to concentration by dividing by the volume (2 L): $$[\mathrm{C_{6}H_{5}CHO}] = \frac{x}{2} = \frac{0.00832~\mathrm{mol}}{2~\mathrm{L}} = 0.00416~\mathrm{M}$$ Now, using the ideal gas law, we can convert the molar concentration to partial pressure: $$P_{\mathrm{C_{6}H_{5}CHO}} = [\mathrm{C_{6}H_{5}CHO}] RT = (0.00416~\mathrm{M}) (0.0821~\mathrm{L \cdot atm / (mol \cdot K)}) (250+273.15)$$ $$P_{\mathrm{C_{6}H_{5}CHO}} \approx 0.74~\mathrm{atm}$$ The partial pressure of benzaldehyde at equilibrium is approximately 0.74 atm.
06

Determine the mass of benzyl alcohol at equilibrium

Use the equilibrium moles of benzyl alcohol from the ICE table and convert to mass using the molar mass: $$\mathrm{Mass~of~benzyl~alcohol~at~equilibrium} = (0.0139 - x) \cdot 108.14~\mathrm{g/mol}$$ $$\mathrm{Mass~of~benzyl~alcohol~at~equilibrium} = (0.0139 - 0.00832) \cdot 108.14~\mathrm{g/mol} \approx 0.605~\mathrm{g}$$ There are approximately 0.605 grams of benzyl alcohol remaining at equilibrium.

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Most popular questions from this chapter

Consider the system $$\mathrm{A}(g)+2 \mathrm{~B}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{D}(g)$$ at \(25^{\circ} \mathrm{C}\). At zero time, only \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) are present. The reaction reaches equilibrium \(10 \mathrm{~min}\) after the reaction is initiated. Partial pressures of \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{D}\) are written as \(P_{\mathrm{A}}, P_{\mathrm{B}}\), and \(P_{\mathrm{D}}\). Answer the questions below, using LT (for is less than), GT (for is greater than), EQ (for is equal to), or MI (for more information required). (a) \(P_{\mathrm{D}}\) at \(11 \mathrm{~min}\) ________ \(P_{\mathrm{D}}\) at \(12 \mathrm{~min} .\) (b) \(P_{\mathrm{A}}\) at \(5 \mathrm{~min}\) \(P_{\mathrm{A}}\) ______ at \(7 \mathrm{~min}\) (c) \(K\) for the forward reaction ______ \(K\) for the reverse reaction. (d) At equilibrium, \(K\)______Q. (e) After the system is at equilibrium, more of gas \(\mathrm{B}\) is added. After the system returns to equilibrium, \(K\) before the addition of \(B\) \(K\) _____ after the addition of \(\mathrm{B}\). (f) The same reaction is initiated, this time with a catalyst. \(K\) for the system without a catalyst _____ \(K\) for the system with a catalyst. (g) \(K\) for the formation of one mole of \(\mathrm{D}\) \(K\) _____ for the formation of two moles of \(\mathrm{D}\). (h) The temperature of the system is increased to \(35^{\circ} \mathrm{C} . P_{\mathrm{B}}\) at equilibrium at \(25^{\circ} \mathrm{C} \longrightarrow P_{\mathrm{B}}\) _______at equilibrium at \(35^{\circ} \mathrm{C}\). (i) Ten more grams of \(\mathrm{C}\) are added to the system. \(P_{\mathrm{B}}\) before the addition of \(\mathrm{C} \quad P_{\mathrm{B}}\) _____ after the addition of \(\mathrm{C}\).

Isopropyl alcohol is the main ingredient in rubbing alcohol. It can decompose into acetone (the main ingredient in nail polish remover) and hydrogen gas according to the following reaction: $$\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{CO}(g)+\mathrm{H}_{2}(g)$$ At \(180^{\circ} \mathrm{C}\), the equilibrium constant for the decomposition is \(0.45\). If \(20.0 \mathrm{~mL}\) \((d=0.785 \mathrm{~g} / \mathrm{mL})\) of isopropyl alcohol is placed in a \(5.00\) -L vessel and heated to \(180^{\circ} \mathrm{C}\), what percentage remains undissociated 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 \(1000 \mathrm{~K}\).

A 1.0-L reaction vessel at \(90^{\circ} \mathrm{C}\) contains \(8.00 \mathrm{~g}\) of sulfur, hydrogen, and hydrogen sulfide gases with partial pressures of \(6.0 \mathrm{~atm}\) and \(0.40 \mathrm{~atm}\), respectively, at equilibrium: $$\mathrm{H}_{2}(g)+\mathrm{S}(s) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)$$ (a) Calculate \(K\) for the reaction at equilibrium. (b) The mass of sulfur is increased to \(10.0\) grams. What are the partial pressures of \(\mathrm{H}_{2}\) and \(\mathrm{H}_{2} \mathrm{~S}\) when equilibrium is reestablished? (c) The pressure of \(\mathrm{H}_{2} \mathrm{~S}\) is increased to \(1.0 \mathrm{~atm}\). What are the partial pressures of \(\mathrm{H}_{2}\) and \(\mathrm{H}_{2} \mathrm{~S}\) when equilibrium is reestablished?

The reaction $$\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$$ has an equilibrium constant of \(1.30\) at \(650^{\circ} \mathrm{C}\). Carbon monoxide and steam both have initial partial pressures of \(0.485 \mathrm{~atm}\), while hydrogen and carbon dioxide start with partial pressures of \(0.159\) atm. (a) Calculate the partial pressure of each gas at equilibrium. (b) Compare the total pressure initially with the total pressure at equilibrium. Would that relation be true of all gaseous systems?
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