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

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 $NaHCO_{3}$ = $5 g *\frac{1 mol}{84 g} = 0.05952 mol NaHCO_{3}$

Volume of HCl solution = $50 cm^{3} * \frac{1 mL}{1cm^{3}} = 50 mL$

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$\frac{J}{g. ^{0}C}$

ΔT = Temperature difference = 6.8 K = (6.8 -273 ) C = -266.2$^{0}C$

$Q = 55 g * 4.184 \frac{J}{g. K}(6.8K) = 1565 J$

Enthalpy of neutralization per mole of $NaHCO_{3}$

= $\frac{1565J}{0.05952 mol} = 26294 \frac{J}{mol}*\frac{1 kJ}{1000J} =26.294kJ/mol$

Answer 2

Answer:

23.9 kJ/mol

Explanation:

Answer for Founder's Education/Educere