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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

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 $$\mathrm{C}(s)+\mathrm{CO}_{2}(g)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{COCl}_{2}(g)$$

When carbon monoxide reacts with hydrogen gas, methane and steam are formed. $$\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ At \(1127^{\circ} \mathrm{C}\), analysis at equilibrium shows that \(P_{\mathrm{CO}}=0.921 \mathrm{~atm}\), \(P_{\mathrm{H}_{2}}=1.21 \mathrm{~atm}, P_{\mathrm{CH}_{4}}=0.0391 \mathrm{~atm}\), and \(P_{\mathrm{H}_{2} \mathrm{O}}=0.0124 \mathrm{~atm} .\) What is the equilibrium constant, \(K\), for the reaction at \(1127^{\circ} \mathrm{C}\) ?

Consider the decomposition of ammonium hydrogen sulfide: $$\mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g)$$ In a sealed flask at \(25^{\circ} \mathrm{C}\) are \(10.0 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{HS}\), ammonia with a partial pressure of \(0.692 \mathrm{~atm}\), and \(\mathrm{H}_{2} \mathrm{~S}\) with a partial pressure of \(0.0532 \mathrm{~atm}\). When equilibrium is established, it is found that the partial pressure of ammonia has increased by 12.4\%. Calculate \(K\) for the decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) at \(25^{\circ} \mathrm{C}\).

Predict the direction in which each of the following equilibria will shift if the pressure on the system is increased by compression. (a) \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{C}(s) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) (b) \(\mathrm{SbCl}_{3}(g) \rightleftharpoons \mathrm{SbCl}_{3}(g)+\mathrm{Cl}_{2}(g)\) (c) \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\)

Consider the system \(4 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-1530.4 \mathrm{~kJ}\) (a) How will the amount of ammonia at equilibrium be affected by (1) removing \(\mathrm{O}_{2}(g)\) ? (2) adding \(\mathrm{N}_{2}(g)\) ? (3) adding water? (4) expanding the container at constant pressure? (5) increasing the temperature? (b) Which of the above factors will increase the value of \(K ?\) Which will decrease it?
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