Respuesta :
Answer: The calculated molar enthalpy change of reaction for his second trial was -108 kJ.
Explanation:-
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
[tex]\text{no of moles}={\text{Molarity}\times {\text{Volume in L}}[/tex]
Thus [tex]\text{no of moles}of HCl={1.0M}\times {0.05L}=0.05moles[/tex]
Thus [tex]\text{no of moles}of NaOH={1.0M}\times {0.05L}=0.05moles[/tex]
[tex]HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)[/tex]
Given for second trial:
[tex]\text{no of moles}of HCl={1.0M}\times {0.1L}=0.1moles[/tex]
[tex]\text{no of moles}of NaOH={1.0M}\times {0.1L}=0.1moles[/tex]
0.05 moles of [tex]HCl[/tex] reacts with 0.05 moles of [tex]NaOH[/tex] to release heat = 54 kJ
0.1 moles of [tex]HCl[/tex] reacts with 0.05 moles of [tex]NaOH[/tex] to release heat =[tex]\frac{54}{0.05}\times 0.1=108kJ[/tex]
Thus calculated molar enthalpy change of reaction for his second trial was -108 kJ.
