Chemistry, 11.11.2020 18:50 290000992

Please answer I have a finals and I need help with a question similar!!! 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 °C. 345 mL of hydrogen gas at 7.12 bar and 271 K is introduced. Assume 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 to22.0 °C?. units in bar please

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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

The partial pressure of methanol is $P_{CH_3OH_{(g)}} =0.077 \ bar$

The partial pressure of carbon monoxide is $P_{CO} = 0.382 \ bar$

The partial pressure at hydrogen is $P_H = O \ bar$

Explanation:

From the question we are told that

The volume of the flask is $V_f = 10.0 \ L$

The initial pressure of carbon monoxide gas is $P_{CO} = 0.461 \ bar$

The initial temperature of carbon monoxide gas is $T_{CO} = 22.0^oC$

The volume of the hydrogen gas is $V_h = 200 mL = 200 *10^{-3} \ L$

The initial pressure of the hydrogen is $P_H = 7.10 \ bar$

The initial temperature of the hydrogen is $T_H = 271 \ K$

The reaction of carbon monoxide and hydrogen is represented as

$CO_{(g)} + 2H_2_{(g)} \rightarrow CH_3OH_{(g)}$

Generally from the ideal gas equation the initial number of moles of carbon monoxide is

$n_1 = \frac{P_{CO} * V_f }{RT_{CO}}$

Here R is the gas constant with value $R = 0.0821 \ L \cdot atm \cdot mol^{-1} \cdot K$

=> $n_1 = \frac{0.461 * 10 }{0.0821 * (22 + 273)}$

=> n_1 = 0.19

Generally from the ideal gas equation the initial number of moles of Hydrogen is

$n_2 = \frac{P_{H} * V_H }{RT_{H}}$

$n_2 = \frac{ 7.10 * 0.2 }{0.0821 * 271 }$

=> n_2 = 0.064

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

$x = \frac{0.064}{2}$

=> $x = 0.032 \ moles \ of \ CO$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of $CH_3OH_{(g)}$

=> 0.064 moles of hydrogen gas will react with z mole of $CH_3OH_{(g)}$

$z = \frac{0.064}{2}$

=> $z = 0.032 \ moles \ of \ CH_3OH_{(g)}$

From this calculation we see that the limiting reactant is hydrogen

Hence the remaining CO after the reaction is

n_k = n_1 - x

=> n_k = 0.19 - 0.032

=> n_k = 0.156

So at the end of the reaction , the partial pressure for CO is mathematically represented as

$P_{CO} = \frac{n_k * R * T_{CO}}{V}$

=> $P_{CO} = \frac{0.158 * 0.0821 * 295}{10}$

=> $P_{CO} = 0.382 \ bar$

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

$P_{CH_3OH_{(g)}} = \frac{z * R * T_{CO}}{V_f}$

Here $T_{CO}$ is used because it is given the question that the temperature returned to 22.0 degrees C

$P_{CH_3OH_{(g)}} = \frac{0.03 * 0.0821 * 295}{10}$

$P_{CH_3OH_{(g)}} =0.077 \ bar$

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