Question

In a typical reaction, the following results were obtained: Part A: 26 mL of 1.069 M HCl were neutralized with 24 mL of 1.2 M NH3. The initial temperature was 23.0 °C, the final temperature was 30.9 °C. Part B: 2.205 g of NH4Cl were dissolved in 50 mL of water. The initial temperature was 23.0 °C, the final temperature was 21.0 °C. Calculate ΔH5.

Answer 1

part A)

moles of HCl = 26 x 1.069 / 1000 = 0.0278

moles of NH3 = 24 x 1.2 / 1000 = 0.0288

NH3 +   HCl   ----------------> NH4Cl

1            1

0.0288   0.0278

here limitng reagent is HCl

mass of solution = 26 + 24 = 50 g

Q = m Cp dT

   = 50 x 4.184 x (30.9 - 23)

   = 1652.68 J

ΔH = - Q /n = - 1652.68 x 10^-3 / 0.02779

ΔH = - 59.46 kJ/mol

part B)

Q = m Cp dT

   = 50 x 4.184 x (21 - 23)

= - 418.4 J

moles of NH4Cl = 2.205 / 53.49 = 0.0412

ΔH = - Q /n = 418.4 x 10^-3 / 0.0412

ΔH = 10.15 kJ/mol