Complete Question
methanol can be synthesized in the gas phase by the reaction of gas phase carbon monoxide with gas phase hydrogen, a 10.0 L reaction flask contains carbon monoxide gas at 0.461 bar and 22.0 degrees Celsius. 200 mL of hydrogen gas at 7.10 bar and 271 K is introduced. Assuming the reaction goes to completion (100% yield)
what are the partial pressures of each gas at the end of the reaction, once the temperature has returned to 22.0 degrees C express final answer in units of bar
Answer:
The partial pressure of methanol is ![P_{CH_3OH_{(g)}} =0.077 \ bar](https://tex.z-dn.net/?f=P_%7BCH_3OH_%7B%28g%29%7D%7D%20%3D0.077%20%5C%20%20bar)
The partial pressure of carbon monoxide is ![P_{CO} = 0.382 \ bar](https://tex.z-dn.net/?f=P_%7BCO%7D%20%3D%200.382%20%5C%20bar)
The partial pressure at hydrogen is ![P_H = O \ bar](https://tex.z-dn.net/?f=P_H%20%3D%20%20O%20%5C%20%20bar)
Explanation:
From the question we are told that
The volume of the flask is ![V_f = 10.0 \ L](https://tex.z-dn.net/?f=V_f%20%3D%2010.0%20%5C%20%20L)
The initial pressure of carbon monoxide gas is ![P_{CO} = 0.461 \ bar](https://tex.z-dn.net/?f=P_%7BCO%7D%20%3D%200.461%20%5C%20bar)
The initial temperature of carbon monoxide gas is ![T_{CO} = 22.0^oC](https://tex.z-dn.net/?f=T_%7BCO%7D%20%3D%2022.0%5EoC)
The volume of the hydrogen gas is ![V_h = 200 mL = 200 *10^{-3} \ L](https://tex.z-dn.net/?f=V_h%20%20%3D%20%20200%20mL%20%3D%20200%20%2A10%5E%7B-3%7D%20%5C%20%20L)
The initial pressure of the hydrogen is ![P_H = 7.10 \ bar](https://tex.z-dn.net/?f=P_H%20%20%3D%20%207.10%20%5C%20%20bar)
The initial temperature of the hydrogen is ![T_H = 271 \ K](https://tex.z-dn.net/?f=T_H%20%3D%20271%20%5C%20%20K)
The reaction of carbon monoxide and hydrogen is represented as
![CO_{(g)} + 2H_2_{(g)} \rightarrow CH_3OH_{(g)}](https://tex.z-dn.net/?f=CO_%7B%28g%29%7D%20%2B%202H_2_%7B%28g%29%7D%20%5Crightarrow%20CH_3OH_%7B%28g%29%7D)
Generally from the ideal gas equation the initial number of moles of carbon monoxide is
![n_1 = \frac{P_{CO} * V_f }{RT_{CO}}](https://tex.z-dn.net/?f=n_1%20%20%3D%20%20%5Cfrac%7BP_%7BCO%7D%20%2A%20%20V_f%20%7D%7BRT_%7BCO%7D%7D)
Here R is the gas constant with value ![R = 0.0821 \ L \cdot atm \cdot mol^{-1} \cdot K](https://tex.z-dn.net/?f=R%20%20%3D%200.0821%20%5C%20L%20%5Ccdot%20atm%20%5Ccdot%20mol%5E%7B-1%7D%20%5Ccdot%20K)
=>
=>
Generally from the ideal gas equation the initial number of moles of Hydrogen is
![n_2 = \frac{P_{H} * V_H }{RT_{H}}](https://tex.z-dn.net/?f=n_2%20%20%3D%20%20%5Cfrac%7BP_%7BH%7D%20%2A%20%20V_H%20%7D%7BRT_%7BH%7D%7D)
![n_2 = \frac{ 7.10 * 0.2 }{0.0821 * 271 }](https://tex.z-dn.net/?f=n_2%20%20%3D%20%20%5Cfrac%7B%207.10%20%2A%20%200.2%20%7D%7B0.0821%20%2A%20271%20%7D)
=> ![n_2 = 0.064](https://tex.z-dn.net/?f=n_2%20%20%3D%20%200.064)
Generally from the chemical equation of the reaction we see that
2 moles of hydrogen gas reacts with 1 mole of CO
=> 0.064 moles of hydrogen gas will react with x mole of CO
So
![x = \frac{0.064}{2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B0.064%7D%7B2%7D)
=> ![x = 0.032 \ moles \ of \ CO](https://tex.z-dn.net/?f=x%20%3D%200.032%20%5C%20moles%20%5C%20of%20%20%5C%20%20CO)
Generally from the chemical equation of the reaction we see that
2 moles of hydrogen gas reacts with 1 mole of ![CH_3OH_{(g)}](https://tex.z-dn.net/?f=CH_3OH_%7B%28g%29%7D)
=> 0.064 moles of hydrogen gas will react with z mole of ![CH_3OH_{(g)}](https://tex.z-dn.net/?f=CH_3OH_%7B%28g%29%7D)
So
![z = \frac{0.064}{2}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B0.064%7D%7B2%7D)
=> ![z = 0.032 \ moles \ of \ CH_3OH_{(g)}](https://tex.z-dn.net/?f=z%20%3D%200.032%20%5C%20moles%20%5C%20of%20%20%5C%20CH_3OH_%7B%28g%29%7D)
From this calculation we see that the limiting reactant is hydrogen
Hence the remaining CO after the reaction is
![n_k = n_1 - x](https://tex.z-dn.net/?f=n_k%20%3D%20n_1%20-%20x)
=> ![n_k = 0.19 - 0.032](https://tex.z-dn.net/?f=n_k%20%3D%200.19%20%20-%200.032)
=> ![n_k = 0.156](https://tex.z-dn.net/?f=n_k%20%3D%200.156)
So at the end of the reaction , the partial pressure for CO is mathematically represented as
![P_{CO} = \frac{n_k * R * T_{CO}}{V}](https://tex.z-dn.net/?f=P_%7BCO%7D%20%3D%20%5Cfrac%7Bn_k%20%20%2A%20%20R%20%2A%20%20T_%7BCO%7D%7D%7BV%7D)
=> ![P_{CO} = \frac{0.158 * 0.0821 * 295}{10}](https://tex.z-dn.net/?f=P_%7BCO%7D%20%3D%20%5Cfrac%7B0.158%20%20%20%2A%20%200.0821%20%2A%20%20295%7D%7B10%7D)
=> ![P_{CO} = 0.382 \ bar](https://tex.z-dn.net/?f=P_%7BCO%7D%20%3D%200.382%20%5C%20bar)
Generally the partial pressure of hydrogen is 0 bar because hydrogen was completely consumed given that it was the limiting reactant
Generally the partial pressure of the methanol is mathematically represented as
![P_{CH_3OH_{(g)}} = \frac{z * R * T_{CO}}{V_f}](https://tex.z-dn.net/?f=P_%7BCH_3OH_%7B%28g%29%7D%7D%20%3D%20%5Cfrac%7Bz%20%20%2A%20%20R%20%2A%20%20T_%7BCO%7D%7D%7BV_f%7D)
Here
is used because it is given the question that the temperature returned to 22.0 degrees C
So
![P_{CH_3OH_{(g)}} = \frac{0.03 * 0.0821 * 295}{10}](https://tex.z-dn.net/?f=P_%7BCH_3OH_%7B%28g%29%7D%7D%20%3D%20%5Cfrac%7B0.03%20%2A%200.0821%20%2A%20%20295%7D%7B10%7D)
![P_{CH_3OH_{(g)}} =0.077 \ bar](https://tex.z-dn.net/?f=P_%7BCH_3OH_%7B%28g%29%7D%7D%20%3D0.077%20%5C%20%20bar)