Find the enthalpy of neutralization of HCl and NaOH. 87 cm3 of 1.6 mol dm-3 hydrochloric acid was neutralized by 87 cm3 of 1.6 mol dm-3 NaOH. The temperature rose from 298 K to 317.4 K. The specific heat capacity is the same as water, 4.18 J/K g.

A. - 101.37 kJ

B. 7055 kJ

C. 10,1365 kJ

The correct answer is A : - 101.37 KJ

Explanation:

Specific heat, s = 4.18 J/Kg

Density of water is 1g/cm3, so 174 cm3 of total solution is = 174 g

Total mass of reaction mixture is = 174 g

Rise in tempetrature, ΔT = 317.4 K - 298 K = 19.4 K

87 cm3 of 1.6 mol dm-3 of HCl = 87 cm3 of 1.6 mol dm-3 NaOH

1.6 M solution means that 1000 cm3 of solution has 1.6 moles.

So, 87 cm3 of 1.6 M solutions = 0.1392 M of HCl and NaOH

Heat evolved (q) = m x s x ΔT

= 174 g x 4.18 J/Kg X 19.4 K

=14110.0 J = 14.110 KJ (for exothermirmic reaction - 14.110 KJ)

Enthalpy of neutralization = - 14.110 KJ / 0.1392 = - 101.37 KJ