Respuesta :
Answer : The entropy change of reaction for 1.62 moles of [tex]BrF_3[/tex] reacts at standard condition is 217.68 J/K
Explanation :
The given balanced reaction is,
[tex]2BrF_3(g)\rightarrow Br_2(g)+3F_2(g)[/tex]
The expression used for entropy change of reaction [tex](\Delta S^o)[/tex] is:
[tex]\Delta S^o=S_f_{product}-S_f_{reactant}[/tex]
[tex]\Delta S^o=[n_{Br_2}\times \Delta S_f^0_{(Br_2)}+n_{F_2}\times \Delta S_f^0_{(F_2)}]-[n_{BrF_3}\times \Delta S_f^0_{(BrF_3)}][/tex]
where,
[tex]\Delta S^o[/tex] = entropy change of reaction = ?
n = number of moles
[tex]\Delta S_f^0[/tex] = standard entropy of formation
[tex]\Delta S_f^0_{(Br_2)}[/tex] = 245.463 J/mol.K
[tex]\Delta S_f^0_{(F_2)}[/tex] = 202.78 J/mol.K
[tex]\Delta S_f^0_{(BrF_3)}[/tex] = 292.53 J/mol.K
Now put all the given values in this expression, we get:
[tex]\Delta S^o=[1mole\times (245.463J/K.mole)+3mole\times (202.78J/K.mole)}]-[2mole\times (292.53J/K.mole)][/tex]
[tex]\Delta S^o=268.74J/K[/tex]
Now we have to calculate the entropy change of reaction for 1.62 moles of [tex]BrF_3[/tex] reacts at standard condition.
From the reaction we conclude that,
As, 2 moles of [tex]BrF_3[/tex] has entropy change = 268.74 J/K
So, 1.62 moles of [tex]BrF_3[/tex] has entropy change = [tex]\frac{1.62}{2}\times 268.74=217.68J/K[/tex]
Therefore, the entropy change of reaction for 1.62 moles of [tex]BrF_3[/tex] reacts at standard condition is 217.68 J/K
