Can Someone please help me with this problem? Question You want to determine AH

Question

Can Someone please help me with this problem?

Question You want to determine AH for the reaction Zn(s) 2HCl(aa) ZnCl2(ag) H20g) To do so, you first determine the heat capacity of a calorimeter using the following reaction, whose AH is known: NaOH(aq) HCl(aq) Naci(aa) H20 (l) AH 57.32 kJ (a) Calculate the heat capacity of the calorimeter from these data: Amounts used: 50.0 mL of 2.00 M HCl and 50.0 mL of 2.00 M NaOH Initial T of both solutions: 16.9oC Maximum T recorded during reaction: 30.4°C Density of resulting NaCl solution: 1.04 g/mL c of 1.00 M Naci(ag): 3.93 J/g K

Explanation / Answer

a) heat generated in reaction = heat used to raise temp of water + heat used to raise temp of calorimeter

= specific heat of solution x mass of solution x temp change + specific heat of calorimeter x temp change

sol vol = 100 ml , soln mass = vol x density = 100 x 1.04 = 104 gm , temp change = 30.4-16.9 = 13.5 ,

heat generated in rxn = moles of HCl ( or NaOH) reacted x delta H = ( 2 x 50/1000) x 57.32 = 5.732 ( -ve sign removed as we mentioned heat generated)

now final eq 5.732 = (3.93 x ( 104) x 13.5 ) /1000 + C x 13.5

C = 0.0159 KJ/C

b) dT = 20.8-16.8 = 4 , HCl mass = 100 x 1.015 = 101.5 , Zn mass = 1.3078 , total mass = 101.5+1.3078 = 102.8078

heat released in reaction = heat used by Zn soln to gain temp + heat used by calorimter to gain temp

heat released = (102.8078 x 3.95 x 4/1000) + 0.0159 x 4   

= 1.688 KJ

moles of Zn used = 1.3078/65.4 = 0.02

dH of reaction per mole = 1.688/0.02 = -84.4 KJ/mol ( -ve sign indicates heat released)

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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