Respuesta :
Answer : The entropy change for the surroundings when 1.74 moles of Na(s) react at standard conditions is [tex]4.38\times 10^4J/g.K[/tex]
Explanation :
The given balanced chemical reaction is,
[tex]2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq)+H_2(g)[/tex]
First we have to calculate the enthalpy of reaction [tex](\Delta H^o)[/tex].
[tex]\Delta H^o=H_f_{product}-H_f_{reactant}[/tex]
[tex]\Delta H^o=[n_{NaOH}\times \Delta H_f^0_{(NaOH)}+n_{H_2}\times \Delta H_f^0_{(H_2)}]-[n_{Na}\times \Delta H_f^0_{(Na)+n_{H_2O}\times \Delta H_f^0_{(H_2O)}][/tex]
where,
We are given:
[tex]\Delta H^o_f_{(NaOH(aq))}=-469.15kJ/mol\\\Delta H^o_f_{(H_2(g))}=0kJ/mol\\\Delta H^o_f_{(Na(s))}=0kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(2\times -469.15)+(1\times 0)]-[(2\times 0)+(2\times -285.8)]=-652.5kJ[/tex]
Now we have to calculate the entropy change for surrounding [tex](\Delta S)[/tex].
[tex]\Delta S=\frac{\Delta H^o_{rxn}}{T}[/tex]
where,
[tex]\Delta S[/tex] = change in entropy
[tex]\Delta H^o_{rxn}[/tex] = change in enthalpy = -652.5 kJ
T = temperature = 298 K
Now put all the given values in the above formula, we get:
[tex]\Delta S=\frac{\Delta H^o_{rxn}}{T}[/tex]
[tex]\Delta S=\frac{-652.5kJ}{298K}[/tex]
[tex]\Delta S=\frac{-652.5\times 10^3J}{298K}[/tex]
[tex]\Delta S=-2189.59J/K=-2.19\times 10^3J/K[/tex]
Now we have to calculate the entropy change for the surroundings when 1.74 moles of Na(s) react at standard conditions.
As, 2 moles of Na(s) has entropy change = [tex]-2.19\times 10^3J/K[/tex]
And, 1.74 moles of Na(s) has entropy change = [tex]\frac{1.74mol}{2mol}\times (-2.19\times 10^3J/K)[/tex]
Thus, 23 g of Na(s) has entropy change = [tex]\frac{1.74mol}{2mol}\times (-2.19\times 10^3J/K)\times 23g=4.38\times 10^4J/g.K[/tex]
Therefore, the entropy change for the surroundings when 1.74 moles of Na(s) react at standard conditions is [tex]4.38\times 10^4J/g.K[/tex]
