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AleksAgata [21]
3 years ago
8

Be sure to answer all parts. one of the most important industrial sources of ethanol is the reaction of steam with ethene derive

d from crude oil: c2h4(g) + h2o(g) ⇌ c2h5oh(g)δh o rxn = −47.8 kjkc = 9.00 × 103 at 600. k at equilibrium, the partial pressure of ethanol (c2h5oh) is 200. atm and the partial pressure of water is 400. atm. calculate the partial pressure of ethene (c2h4).
Chemistry
1 answer:
lions [1.4K]3 years ago
7 0

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}]}

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}

[H_{2}O] = \frac{400 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 8.12 \frac{mol}{L}

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}

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

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.

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