Respuesta :
Answer: Temperature for the given reaction is 1040 K (approx).
Explanation:
Formula for enthalpy change of a reaction is as follows.
[tex]\Delta H_{rxn} = \Delta H_{products} - \Delta H_{reactants}[/tex]
For the given reaction equation,
[tex]\Delta H_{rxn} = 2 \times \Delta H_{HI} - (\Delta H_{H_{2}} + \Delta H_{I_{2}})[/tex]
Now, putting the given values into the above formula as follows.
[tex]\Delta H_{rxn} = 2 \times \Delta H_{HI} - (\Delta H_{H_{2}} + \Delta H_{I_{2}})[/tex]
[tex]\Delta H_{rxn} = 2 \times 25.9 - (0 + 62.26)[/tex]
= -10460 J/mol
Now, we will calculate the change in its entropy as follows.
[tex]\Delta S = S_{products} - S_{reactants}[/tex]
= [tex]2 \times S_{HI} - (S_{H_{2}} + S_{I_{2}})[/tex]
= [tex]2 \times 206.3 - (131.0 + 260.6)[/tex]
= 21 J/mol
Also, we know that
[tex]\Delta G = RT ln K_{p} = \Delta H_{rxn} - T\Delta S_{rxn}[/tex]
[tex]-8.314 \times T \times ln(42) = -10460 - T \times 21[/tex]
T = 1036.7 K
= 1040 K
Therefore, we can conclude that temperature for the given reaction is 1040 K (approx).
