Question

5 grams of NaHCO3 was added to 50 cm3 of 1.5 mol dm-3 hydrochloric acid, causing the temperature to decrease by 6.8 K. What is the enthalpy of neutralization per mole of NaHCO3?

Answer 1

23.9 kJ/mol

Step-by-step explanation:

Answer for Founder's Education/Educere

Answer 2

The chemical equation representing the neutralization reaction between HCl and
`NaHCO_(3)` is,

`HCl (aq) + NaHCO_(3)(aq)--> NaCl (aq) + H_(2)O(l)+CO_(2) (g)`

Given mass of
`NaHCO_(3)` = 5g

Moles of
`NaHCO_(3)` =
`5 g *(1 mol)/(84 g) = 0.05952 mol NaHCO_(3)`

Volume of HCl solution =
`50 cm^(3) * (1 mL)/(1cm^(3)) = 50 mL`

Assuming the density of solution to be 1.0 g/mL

Mass of HCl solution = 50 g

Total mass of solution = 50 g+ 5 g = 55 g

Calculating the heat of neutralization:

Q = m C ΔT

m is mass of solution = 55 g

C is the specific heat capacity of the solution = 4.184
`(J)/(g. ^(0)C)`

ΔT = Temperature difference = 6.8 K = (6.8 -273 ) C = -266.2
`^(0)C`

`Q = 55 g * 4.184 (J)/(g. K)(6.8K) = 1565 J`

Enthalpy of neutralization per mole of
`NaHCO_(3)`

=
`(1565J)/(0.05952 mol) = 26294 (J)/(mol)*(1 kJ)/(1000J) =26.294kJ/mol`