Respuesta :
Answer:
The change in entropy of the surrounding is -146.11 J/K.
Explanation:
Enthalpy of formation of iodine gas = [tex]\Delta H_f_{(I_2)}=62.438 kJ/mol[/tex]
Enthalpy of formation of chlorine gas = [tex]\Delta H_f_{(Cl_2)}=0 kJ/mol[/tex]
Enthalpy of formation of ICl gas = [tex]\Delta H_f_{(ICl)}=17.78 kJ/mol[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
For the given chemical reaction:
[tex]I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})][/tex]
[tex]=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol[/tex]
Enthaply change when 1.62 moles of iodine gas recast:
[tex]\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ[/tex]
Entropy of the surrounding = [tex]\Delta S^o_{surr}=\frac{\Delta H}{T}[/tex]
[tex]=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K[/tex]
1 kJ = 1000 J
The change in entropy of the surrounding is -146.11 J/K.
