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GrogVix [38]
3 years ago
13

Equal masses of carbon dioxide (CO2) and oxygen (O2) are mixed. What is the mole

Chemistry
1 answer:
svetoff [14.1K]3 years ago
6 0

The mole fraction of oxygen in this mixture is = 0.5789  which is not listed  in the options. while the mole fraction of carbon dioxide = 0.4211 ( A )

Given that :

Mass of carbon dioxide = mass of oxygen

lets assume the given mass  = x

<u>first step : determine the moles of CO₂ and O₂</u>

moles of CO₂ = mass / molar mass of CO₂

                       = x / 44

moles of O₂  =  mass / molar mass of O2

                     = x / 32

∴ mole fraction of O₂ = moles of O₂ / ( Total moles in the mixture )

                                   = \frac{x}{32} / ( \frac{x}{44} + \frac{x}{32} )

                                   = \frac{44*32*(x)}{32 * 76 *(x)}  = 0.5789

While the mole fraction of Carbon dioxide = 1 - 0.5789 = 0.4211

hence the mole fraction of oxygen in this mixture = 0.5789 while the mole fraction of Carbon dioxide = 0.4211

learn more here : brainly.com/question/15063496

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Answer:

The reactions free energy \Delta G = -49.36 kJ

Explanation:

From the question we are told that

      The pressure of (NO) is P_{NO} = 9.20 \ atm

      The  pressure of  (Cl) gas is  P_{Cl} = 9.15 \ atm

       The  pressure of nitrosly chloride (NOCl) is P_{(NOCl)} = 7.70 \ atm

The reaction is

              2NO_{(g)} + Cl_2 (g)    ⇆   2 NOCl_{(g)}

 From the reaction we can  mathematically evaluate the \Delta G^o (Standard state  free energy ) as

                    \Delta G^o = 2 \Delta G^o _{NOCl} -   \Delta G^o _{Cl_2}  - 2 \Delta G^o _{NO}

The Standard state  free energy for NO is  constant with a value  

                 \Delta G^o _{NO} = 86.55 kJ/mol

 The Standard state  free energy for Cl_2 is  constant with a value                  

             \Delta G^o _{Cl_2} = 0kJ/mol

 The Standard state  free energy for NOCl is  constant with a value

         \Delta G^o _{NOCl} =66.1kJ/mol

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        \Delta G^o = 2 * 66.1 - 0 - 2 * 87.6

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The pressure constant is evaluated as

         Q =  \frac{Pressure \ of  \ product }{ Pressure  \ of \ reactant }

Substituting  values  

        Q = \frac{(7.7)^2 }{(9.2)^2 (9.15) } = \frac{59.29}{774.456}

           = 0.0765

The free energy for this reaction is evaluated as

           \Delta  G  =  \Delta  G^o  + RT ln Q

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          T is temperature in K  with a given value of  T = 25+273 = 298 K

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                \Delta  G  = -43 *10^{3} + 8.314 *298 * ln [0.0765]

                       = -43-6.36

                      \Delta G = -49.36 kJ

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