Thermochemistry and Hess\'s Law Results 1. Enthalpy change NoOH HCl or NHr-HC Co

Question

Thermochemistry and Hess's Law Results 1. Enthalpy change NoOH HCl or NHr-HC Concentration of HCI: 2.0 Concentration and identity of base:2 MaoH Trial Volume of acid in cylinder: Hei Before pouring (mL) After pouring (mL) Om L Volume added (mL) Volume of base in cylinder:Alao so.om Before pouring (mL) After pouring (mL) 50- mL Volume added (mL) 4CC) continued... 119 7 Cengage Leaming All Rights Reserved May not be scanned, copiod or duplicated, or posted to a publicly accesible website, in whole or in

Explanation / Answer

To get moles of limiting reactant, first we obtain the balanced neutralization reaction:

NaOH+ HCl -> NaCl + H2O

We then get respective moles with initial volumes and concentrations given to get limiting reactant. As in both cases same volume quantities are being used, we get once the moles, and that should be the limiting reactant moles:

2.0 M = n / 0.05 L

n = 0.1 moles (trial 1)

2.0 M = n / 0.049 L

n = 0.098 moles (trial 2)

The heat released by system can be calculated with the heat exchange principle, where:

q = mCpdeltaT

We'll assume density and Cp for water, so:

Trial 1

q = 100g * 4.18 J/g K * (35.8 - 22.4)K = 5,601.2 J

Trial 1

q = 98g * 4.18 J/g K * (35.7 - 22.1)K = 5,571.104 J

Trial 1 Trial 2 q(system) 5.601 kJ 5.571 kJ Moles of Limiting Reactant 0.1 moles 0.098 moles delta H (kJ/mol) 5.601 kJ / 0.1 moles = 56.01 kJ/mol 5.571 kJ / 0.098 moles = 56.85 kJ/mol Mean delta H (kJ/mol) 56.43 kJ/mol

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.