Question

Find the temperature at which Kp = 42.0 for the reaction

Answer 1

Answer: Temperature for the given reaction is 1040 K (approx).

Explanation:

Formula for enthalpy change of a reaction is as follows.

     $\Delta H_{rxn} = \Delta H_{products} - \Delta H_{reactants}$

For the given reaction equation,

     $\Delta H_{rxn} = 2 \times \Delta H_{HI} - (\Delta H_{H_{2}} + \Delta H_{I_{2}})$  

Now, putting the given values into the above formula as follows.

    $\Delta H_{rxn} = 2 \times \Delta H_{HI} - (\Delta H_{H_{2}} + \Delta H_{I_{2}})$  

   $\Delta H_{rxn} = 2 \times 25.9 - (0 + 62.26)$  

                    = -10460 J/mol

Now, we will calculate the change in its entropy as follows.

           $\Delta S = S_{products} - S_{reactants}$

                       = $2 \times S_{HI} - (S_{H_{2}} + S_{I_{2}})$

                       = $2 \times 206.3 - (131.0 + 260.6)$

                       = 21 J/mol

Also, we know that

         $\Delta G = RT ln K_{p} = \Delta H_{rxn} - T\Delta S_{rxn}$

         $-8.314 \times T \times ln(42) = -10460 - T \times 21$

                     T = 1036.7 K

                        = 1040 K

Therefore, we can conclude that temperature for the given reaction is 1040 K (approx).