Question

A chemist fills a reaction vessel with 9.20 atm nitrogen monoxide (NO) gas, 9.15 atm chlorine (CI) gas, and 7.70 atm nitrosyl chloride (NOCI) gas at a temperature of 25.0°C. Under these conditions, calculate the reaction free energy AG for the following chemical reaction: 2NO(g) +CI (8) - 2NOCI (8) Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule

Answer 1

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

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

Now substituting this into the equation

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

                $= -43 kJ/mol$

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

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

Where R is gas constant with a value  of  $R = 8.314 J / K \cdot mol$

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

   Substituting value

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

                       $= -43-6.36$

                      $\Delta G = -49.36 kJ$