9. Using Hess's Law and experimental data to determine the enthalpy of the decomposition of sodium hydrogencarbonate to sodium carbonate, water and carbon dioxide

Method (for experimental details see method 1.3a1)

These experiments illustrate how to indirectly determine an enthalpy change that cannot be determined by direct experiment.

In this case the thermal decomposition of sodium hydrogencarbonate.

ΔH^θ for 2NaHCO₃(s)  ===>   Na₂CO₃(s)  +  H₂O(l)  +  CO₂(g)

You cannot directly obtain experimental results to determine (1) the enthalpy of thermal decomposition of sodium hydrogen carbonate. However, you can measure the enthalpy change for two other reactions and using Hess's Law calculate the 'unobtainable' by experiment.

The two other reactions involve (2) mixing sodium hydrogencarbonate and (3) sodium carbonate with hydrochloric acid.

The Hess's Law cycle connecting these three reactions is shown below.

ΔHθ1
2NaHCO3(s)     Na2CO3(s)  +  H2O(l)  +  CO2(g)
+ 2HCl(aq) ΔHθ2				+ 2HCl(aq) ΔHθ3
2NaCl(aq)  +  2H2O(l)  +  2CO2(g)
From Hess's Law:  ΔHθ2  =  ΔHθ1  +  ΔHθ3    or     ΔHθ1  =  ΔHθ2  –  ΔHθ3

By the calorimetric methods already described from 3. to 6. you can separately determine the enthalpy of reaction of firstly.

I see little point in repeating the method and observation details.

(a) sodium hydrogencarbonate with hydrochloric acid and then

(b) the reaction of anhydrous sodium carbonate with hydrochloric acid using the calculation methods already described e.g. experiments 4. to 8.

(a) Examples of experimental results:

atomic masses: Na = 23, H = 1, C = 12, O = 16

formula masses: NaHCO₃ = 84   and   Na₂CO₃ = 106

Enthalpy change for reaction 1.

The unknown!

Enthalpy change for reaction 2.

On adding 3.36 g of sodium hydrogen carbonate to 50 cm³ of 1 molar hydrochloric acid (an excess), the maximum temperature fall was found to be 5.20^oC. Assume the density of hydrochloric acid is the same as pure water.

mol HCl = 1.0 x 50/1000 = 0.05 mol (in excess, check the next line and equation, 1:1 mole ratio)

mol NaHCO₃ = 3.36/84 = 0.04 mol

(the limiting reactant on which the calculation must be based)

NaHCO₃(s)  +  HCl(aq)  ===>  NaCl(aq)  +  H₂O(l)  +  CO₂(g)

q = m c ΔT

q = 50 x 4.18 x 5.2 = 10868.8 J,  1.08688 kJ

Scaling up to 1 mol of NaHCO₃: 1.08688 x 1.0/0.04 = 27.17 kJ

Since the temperature fell, the reaction is endothermic, heat absorbed

Therefore ΔH^θ_r,298 = ΔH^θ₂ = +27.2 kJ mol^–1 per mole of NaHCO₃

(equivalent to +54.4 kJmol^-1 for 2 mol of NaHCO₃)

then

(b) On adding 2.12 g of anhydrous sodium carbonate to 50 cm³ of 1 molar hydrochloric acid (an excess), the maximum temperature rise was estimated to be 3.30^oC.

mol HCl = 1.0 x 50/1000 = 0.05 mol (in excess, check the next few lines!)

mol Na₂CO₃ = 2.12/106 = 0.02 mol (the limiting reactant on which the calculation must be based)

Na₂CO₃(s)  +  2HCl(aq)  ===>  2NaCl(aq)  +  H₂O(l)  +  CO₂(g)

0.02 mol Na₂CO₃ will react with 0.04 mol HCl

q = m c ΔT,  q = 50 x 4.18 x 3.3 = 689.7 J,  0.6897 kJ

Scaling up to 1 mol of Na₂CO₃: 1.08688 x 1.0/0.02 = 34.485 kJ

Since the temperature rose, the reaction is exothermic, heat released

Therefore ΔH^θ_r,298 = ΔH^θ₃ = –34.5 kJ mol^–1 per mole of Na₂CO₃

The temperature rises are quite small, particularly for reaction 3., so I would suggest using double the masses and 2 molar hydrochloric acid. However, the real heat capacity of the resulting sodium chloride solution is likely to be <3.9 J g^{–1 o}C^–1.

 

From this data, using a Hess's Law cycle described below, you can then calculate the enthalpy of hydration of anhydrous copper(II) sulfate. Note that reaction 2 has double molar quantities to that the cycle as a whole is perfectly balanced.

ΔH^θ₁  =  ΔH^θ₂  –  ΔH^θ₃

ΔH^θ₁  =  +54.4  –  (–34.5)  =  +88.9 kJ mol^–1

for the thermal decomposition of NaHCO₃

i.e. ΔH^θ_reaction for 2NaHCO₃(s)  ===>   Na₂CO₃(s)  +  H₂O(l)  +  CO₂(g)

After the Hess's Law cycle below (repeat!), for this experiment, I've worked everything out from scratch using accurate thermodynamic date i.e. all the enthalpies of formation to calculate all three enthalpies of reaction.

ΔHθ1
2NaHCO3(s)     Na2CO3(s)  +  H2O(l)  +  CO2(g)
+ 2HCl(aq) ΔHθ2				+ 2HCl(aq) ΔHθ3
2NaCl(aq)  +  2H2O(l)  +  2CO2(g)
From Hess's Law:  ΔHθ2  =  ΔHθ1  +  ΔHθ3    and     ΔHθ1  =  ΔHθ2  –  ΔHθ3

 

Theoretical calculations of ΔH^θ_1–3 from standard enthalpy of formation data

ΔH^θ_r,298 = standard enthalpy of reaction at 298K and 1 atm/101 kPa pressure

i.e. the usual standard conditions all data in kJ/mol

The thermodynamic data for NaCl(aq) and HCl(aq) apply to concentrations of about 1.0 mol dm^–3 (it varies with concentration)

ΔH^θ_f,298 data (all in kJ mol^–1)

ΔH^θ_f,298(NaHCO₃(s)) = –948

ΔH^θ_f,298(Na₂CO₃(s)) = –1131

ΔH^θ_f,298(H₂O(l)) = –286

ΔH^θ_f,298(CO₂(g)) = –394

ΔH^θ_f,298(NaCl(aq, ~1.0 mol dm^–3)) = –407

ΔH^θ_f,298(HCl(aq, ~1.0 mol dm^–3)) = –165

 

Theoretical calculation 1.

(ΔH^θ₁) 2NaHCO₃(s)  ===>   Na₂CO₃(s)  +  H₂O(l)  +  CO₂(g)

ΔH^θ_r,298 = ∑ΔH^θ_f,298(products) – ∑ΔH^θ_f,298(reactants)

ΔH^θ_r,298 = {ΔH^θ_f,298(Na₂CO₃(s))  +  ΔH^θ_f,298(H₂O(l))  +  ΔH^θ_f,298(CO₂(g))} – {2 x ΔH^θ_f,298(NaHCO₃(s))}

ΔH^θ_r,298 = {–1131 +   –286  +  –394}  –  {2 x –948)

ΔH^θ_r,298  = –1811 – (–1896)  = ΔH^θ₁ = +85 kJ mol^–1

The thermal decomposition of sodium hydrogencarbonate is endothermic.

 

Theoretical calculation 2.

(ΔH^θ₂) 2NaHCO₃(s)  +  2HCl(aq)  ===>  2NaCl(aq)  +  2H₂O(l)  +  2CO₂(g)

ΔH^θ_r,298 = ∑ΔH^θ_f,298(products) – ∑ΔH^θ_f,298(reactants)

ΔH^θ_r,298 = {2 x ΔH^θ_f,298(NaCl(aq))  + 2 x ΔH^θ_f,298(H₂O(l))  +  2 x ΔH^θ_f,298(CO₂(g))}  –  {2 x ΔH^θ_f,298(NaHCO₃(s))  +  2 x ΔH^θ_f,298(HCl(aq))}

ΔH^θ_r,298 = {(2 x –407) + (2 x –286) + (2 x –394)} – {(2 x –948) + (2 x –165)}

= –2174 – (–2226) = ΔH^θ₂ = +52 kJ mol^–1

(equivalent to +26 kJ mol^–1 per mole of NaHCO₃)

Therefore the reaction of sodium hydrogen carbonate with hydrochloric acid is endothermic, so you should see a temperature fall in the calorimeter solution.

 

Theoretical calculation 3.

(ΔH^θ₃) Na₂CO₃(s)  +  2HCl(aq)  ===>  2NaCl(aq)  +  H₂O(l)  +  CO₂(g)

ΔH^θ_r,298 = ∑ΔH^θ_f,298(products) – ∑ΔH^θ_f,298(reactants)

ΔH^θ_r,298 =  {2 x ΔH^θ_f,298(NaCl(aq))  + ΔH^θ_f,298(H₂O(l))  +  ΔH^θ_f,298(CO₂(g))} – {ΔH^θ_f,298(Na₂CO₃(s))  +  2 x ΔH^θ_f,298(HCl(aq))}

ΔH^θ_r,298 = {(2 x –407)  +  –286  +  –394} – {–1131  +  (2 x –165)} = –1494 – (–1461)

= ΔH^θ₃ = –33 kJ mol^–1

Therefore the reaction of sodium carbonate with hydrochloric acid is exothermic, so you should see a temperature rise in the calorimeter solution.

 

Using the theoretical data from calculations 2 and 3 to simulate 'perfect' experimental results:

ΔH^θ₁  =  ΔH^θ₂  –  ΔH^θ₃  =  +52 – (–33)  = +85 kJ mol^–1

quite close to the first, partly experimental, which was a great relief after dealing with all those numbers and brackets!

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Experimental methods of determining enthalpy changes

Energetics–Thermochemistry–Thermodynamics Notes INDEX

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