Answer: 2.17x10⁻³ atm
Explanation:
First, we must write the balanced chemical equation for the process:
C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g)
The chemical reactions that occur in a closed container can reach a state of <u>chemical equilibrium</u> that is characterized because the concentrations of the reactants and products remain constant over time. The <u>equilibrium constant</u> of a chemical reaction is the value of its reaction quotient in chemical equilibrium.
The equilibrium constant (K) is expressed as <u>the ratio between the molar concentrations (mol/L) of reactants and products.</u> Its value in a chemical reaction depends on the temperature, so it must always be specified.
<u>We will use the the equilibrium constant Kc of the reaction to calculate partial pressure of ethene.</u> The constant Kc for the above reaction is,
Kc = ![\frac{[C_{2} H_{5}OH]}{[H_{2}O][C_{2} H_{4}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BC_%7B2%7D%20H_%7B5%7DOH%5D%7D%7B%5BH_%7B2%7DO%5D%5BC_%7B2%7D%20H_%7B4%7D%5D%7D)
According to the law of ideal gases,
PV = nRT
where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the gas constant (0.082057 atm L / mol K)
.
We can use the ideal gas law to determine the molar concentrations ([x] = n / V) from the gas pressures of ethanol and water, assuming that all gases involved behave as ideal gases. In this way,
PV = nRT → P = (n/V) RT → P = [x] RT → [x] = P / RT
So,
![[C_{2} H_{5}OH] = \frac{200 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 4.06 \frac{mol}{L}](https://tex.z-dn.net/?f=%5BC_%7B2%7D%20H_%7B5%7DOH%5D%20%3D%20%5Cfrac%7B200%20atm%7D%7B0.082057%20%5Cfrac%7Batm%20L%7D%7Bmol%20K%7D%20x%20600%20K%20%7D%20%3D%204.06%20%5Cfrac%7Bmol%7D%7BL%7D)
![[H_{2}O] = \frac{400 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 8.12 \frac{mol}{L}](https://tex.z-dn.net/?f=%5BH_%7B2%7DO%5D%20%3D%20%5Cfrac%7B400%20atm%7D%7B0.082057%20%5Cfrac%7Batm%20L%7D%7Bmol%20K%7D%20x%20600%20K%20%7D%20%3D%208.12%20%5Cfrac%7Bmol%7D%7BL%7D)
So, the molar concentration of ethene (C₂H₄) will be,
![[C_{2} H_{4}] = \frac{[C_{2} H_{5}OH]}{[H_{2}O] x Kc} = \frac{4.06 \frac{mol}{L} }{8.12 \frac{mol}{L}x9.00 x 10^{3} \frac{L}{mol} } = 5.56 x 10^{-5}\frac{mol}{L}](https://tex.z-dn.net/?f=%5BC_%7B2%7D%20H_%7B4%7D%5D%20%3D%20%5Cfrac%7B%5BC_%7B2%7D%20H_%7B5%7DOH%5D%7D%7B%5BH_%7B2%7DO%5D%20x%20Kc%7D%20%3D%20%5Cfrac%7B4.06%20%5Cfrac%7Bmol%7D%7BL%7D%20%7D%7B8.12%20%5Cfrac%7Bmol%7D%7BL%7Dx9.00%20x%2010%5E%7B3%7D%20%5Cfrac%7BL%7D%7Bmol%7D%20%7D%20%3D%205.56%20x%2010%5E%7B-5%7D%5Cfrac%7Bmol%7D%7BL%7D)
Then, according to the law of ideal gases,
![P_{C_{2} H_{4}} = [C_{2} H_{4}]RT = 5.56 x 10^{-5} \frac{mol}{L} x 0.082057 \frac{atm L}{mol K} x 600 K = 2.17x10^{-3} atm](https://tex.z-dn.net/?f=P_%7BC_%7B2%7D%20H_%7B4%7D%7D%20%3D%20%5BC_%7B2%7D%20H_%7B4%7D%5DRT%20%3D%205.56%20x%2010%5E%7B-5%7D%20%5Cfrac%7Bmol%7D%7BL%7D%20%20x%200.082057%20%5Cfrac%7Batm%20L%7D%7Bmol%20K%7D%20x%20600%20K%20%3D%202.17x10%5E%7B-3%7D%20atm)
So, when the partial pressure of ethanol is 200 atm and the partial pressure of water is 400 atm, the partial pressure of ethene at 600 K is 2.17x10⁻³ atm.
