Question

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 \(2.0-\mathrm{L}\) 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

In summary, for the given reaction at equilibrium, the partial pressure of benzaldehyde is approximately \(0.214\: Atm\), and the amount of benzyl alcohol remaining is approximately \(0.425\:g\).

Step-by-step solution

Convert the given amount of benzyl alcohol into moles.

First, we need to determine the number of moles of benzyl alcohol. To do this, we will use the molar mass of benzyl alcohol, which is approximately \({108.14\:g/mol}\). $$ \text{moles of benzyl alcohol} = \frac{1.50\:g}{108.14\:g/mol} = 0.0139\:\text{mol} $$

Set up the initial and equilibrium concentrations in the reaction.

Next, let's set up the initial and equilibrium concentrations in a table. | | Benzyl Alcohol | Benzaldehyde | Hydrogen Gas | |----------|----------------|--------------|--------------| | Initial | 0.0139 mol | 0 mol | 0 mol | | Change | -x mol | +x mol | +x mol | | Equlbrm | 0.0139-x mol | x mol | x mol | Here, x represents the moles of benzyl alcohol that will reacted to form benzaldehyde and hydrogen gas.

Use the equilibrium constant expression to find the change in the concentrations of the substances and obtain the partial pressure of benzaldehyde.

Now we will use the equilibrium constant expression to solve for x, and then calculate the equilibrium partial pressure of benzaldehyde. The partial pressure of each substance can be calculated using the ideal gas law, \(PV=nRT\). For this problem, the total volume is \(2.0\:L\), and the temperature is \(250^\circ C = 523\:K\). The gas constant R in terms of pressure is \(0.0821 \: Atmosphere \cdot L / mol \cdot K\). $$ K = \frac{P_{\text{benzaldehyde}}\:P_{H_2}}{P_{\text{benzylalcohol}}} \:\Rightarrow\: [0.56] = \frac{x^2/2}{(0.0139-x)/2} $$ Solving for x, we get: $$ x \approx 0.00997\: \text{mol} $$ Now, we can find the partial pressure of benzaldehyde using the formula we derived above: $$ P_{\text{benzaldehyde}} = \frac{nRT}{V} =\frac{(0.00997\:mol)(0.0821\:Atm\:L^1 \cdot mol^{-1} \cdot K^{-1})(523\:K)}{2.0\:L} $$ $$ \Rightarrow\: P_{\text{benzaldehyde}} \approx 0.214\:Atm $$ So, the partial pressure of benzaldehyde is approximately \(0.214\: Atm\)

Calculate the remaining amount of benzyl alcohol in grams.

Now, we will determine the moles of benzyl alcohol remaining at equilibrium, and convert it into grams. From Step 3, we know x is approximately \(0.00997\:\text{mol}\). $$ \text{moles of benzylalcohol at equilibrium} = \text{initial moles - x} = 0.0139 - 0.00997 = 0.00393\: \text{mol} $$ $$ \text{grams of benzylalcohol remaining} = 0.00393 \: \text{mol} \times 108.14 \: \text{g/mol} \approx 0.425\:g $$ Therefore, approximately \(0.425\:g\) of benzyl alcohol remains at equilibrium.