The chemical equation representing the neutralization reaction between HCl and [tex] NaHCO_{3} [/tex] is,
[tex] HCl (aq) + NaHCO_{3}(aq)--> NaCl (aq) + H_{2}O(l)+CO_{2} (g) [/tex]
Given mass of [tex] NaHCO_{3} [/tex] = 5g
Moles of [tex] NaHCO_{3} [/tex] = [tex] 5 g *\frac{1 mol}{84 g} = 0.05952 mol NaHCO_{3} [/tex]
Volume of HCl solution = [tex] 50 cm^{3} * \frac{1 mL}{1cm^{3}} = 50 mL [/tex]
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[tex] \frac{J}{g. ^{0}C} [/tex]
ΔT = Temperature difference = 6.8 K = (6.8 -273 ) C = -266.2[tex] ^{0}C [/tex]
[tex] Q = 55 g * 4.184 \frac{J}{g. K}(6.8K) = 1565 J [/tex]
Enthalpy of neutralization per mole of [tex] NaHCO_{3} [/tex]
= [tex] \frac{1565J}{0.05952 mol} = 26294 \frac{J}{mol}*\frac{1 kJ}{1000J} =26.294kJ/mol [/tex]
