value 10.00 points Sec. Ex. 8-p8 Temperature Change from Reaction 1 out of 10 at

Question

value 10.00 points Sec. Ex. 8-p8 Temperature Change from Reaction 1 out of 10 attempts Ass Enter your answer in the provided box. A quantity of 3.90 x 10 mL of 0.650 M HNO, is mixed with 3.90 x 10 mL of0325 M Ba(OH2 in a constant-pressure calorimeter of negligible heat capacity. The initial S temperature of both solutions is the same at 18.46°C. The heat ofneutralization when 1.00 mol of HNO, reacts with 0.500 mol Ba(oE)2 is -56.2 kJ/mol. Assume that the densities and specific heats of the solution are the same as for water (i.00 g/mLand.184 Jlg OC, respectively). What is the final temperature of the solution?

Explanation / Answer

1.

The reaction between Ba(OH)2 and HNO3 is Ba(OH)2+ 2HNO3-----à Ba(NO3)2+ 2H2O

0.5 moles of Ba(OH)2 reacts with 1mole of HNO3 to generated -56.2 Kj of heat/mole

Moles of HNO3 in 390ml of 0.65M= Molarity* volume (L)= 0.325*650/1000 =0.26 moles

Moles of Ba(OH)2 in 390 ml of 0.325 = 0.325*0.39= 0.13 moles

1 mole of HNO3 produces 56.2 Kj of heat

0.26 moles produced 56.2*0.26 Kj=14.6 KJ

The reaction is exothermic. That is why enthalpy of mixing is –ve.

Volume of solution of HNO3 and Ba(OH)2 = 390+390=780ml

Mass of solution =volume* density= 780 gm

Enthalpy change= mass* specific heat* temperature difference

14.6*1000= 780*4.184* (T- 18.46), T= 22.93 deg.c

T= 22.9 deg.c

2. 1 mole of benzene correspond to 78 gm of benzene ( molar mass of benzene=78 g/mole)

78 gm liberates 49.04 Kj

1 gm produced 49.04/78 Kj/gm =0.63 Kj/gm

Related Questions

Navigate

• Browse (All)
• Browse V
• Subjects

---

---

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