Question

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Answer 1

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  $P_{CH_3OH_{(g)}} =0.077 \ bar$

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

So

          $x = \frac{0.064}{2}$

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

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  $CH_3OH_{(g)}$

So

          $z = \frac{0.064}{2}$

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

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

So

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

     $P_{CH_3OH_{(g)}} =0.077 \ bar$

Answer 2

Answer:

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Explanation: