Respuesta :
Answer : The entropy change of reaction for 2.28 moles of [tex]H_2[/tex] reacts at standard condition is 45.8 J/K
Explanation :
The given balanced reaction is,
[tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(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_{HCl}\times \Delta S_f^0_{(HCl)}]-[n_{H_2}\times \Delta S_f^0_{(H_2)}+n_{Cl_2}\times \Delta S_f^0_{(Cl_2)}][/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_{(H_2)}[/tex] = 130.684 J/mol.K
[tex]\Delta S_f^0_{(Cl_2)}[/tex] = 223.066 J/mol.K
[tex]\Delta S_f^0_{(HCl)}[/tex] = 186.908 J/mol.K
Now put all the given values in this expression, we get:
[tex]\Delta S^o=[2mole\times (186.908J/K.mole)]-[1mole\times (130.684J/K.mole)+1mole\times (223.066J/K.mole)}][/tex]
[tex]\Delta S^o=20.066J/K[/tex]
Now we have to calculate the entropy change of reaction for 2.28 moles of [tex]H_2[/tex] reacts at standard condition.
From the reaction we conclude that,
As, 1 moles of [tex]H_2[/tex] has entropy change = 20.066 J/K
So, 2.28 moles of [tex]H_2[/tex] has entropy change = [tex]\frac{2.28}{1}\times 20.066=45.8J/K[/tex]
Therefore, the entropy change of reaction for 2.28 moles of [tex]H_2[/tex] reacts at standard condition is 45.8 J/K
