In the Solvay process for the production of sodium carbonate (or washing soda), one step is the decomposition of sodium hydrogen carbonate.

Question

2
NaHCO
3
(
𝑠
)
→
Na
2
CO
3
(
𝑠
)
+
CO
2
(
𝑔
)
+
H
2
O
(
𝑔
)
2NaHCO
3
​
(s)→Na
2
​
CO
3
​
(s)+CO
2
​
(g)+H
2
​
O(g)

When 4.3 grams of sodium hydrogen carbonate (NaHCO₃) was placed in a calorimeter and allowed to react completely, 20.0 mL of water surrounding the reaction changed temperature from 29°C to 18.3°C.c) Use Hess’s law to determine the theoretical value of the molar enthalpy change (ΔHdecomp) for NaHCO₃. [4]

Given:
NaHCO
3
(
𝑠
)
+
HCl
(
𝑎
𝑞
)
→
NaCl
(
𝑎
𝑞
)
+
CO
2
(
𝑔
)
+
H
2
O
(
𝑔
)
 
Δ
𝐻
1
=
+
15.85
 
kJ/mol
NaHCO
3
​
(s)+HCl(aq)→NaCl(aq)+CO
2
​
(g)+H
2
​
O(g)ΔH
1
​
=+15.85kJ/mol
Na
2
CO
3
(
𝑠
)
+
2
HCl
(
𝑎
𝑞
)
→
2
NaCl
(
𝑎
𝑞
)
+
CO
2
(
𝑔
)
+
H
2
O
(
𝑔
)
 
Δ
𝐻
2
=
+
13.50
 
kJ/mol
Na
2
​
CO
3
​
(s)+2HCl(aq)→2NaCl(aq)+CO
2
​
(g)+H
2
​
O(g)ΔH
2
​
=+13.50kJ/mol

Answer 1

To determine the theoretical value of the molar enthalpy change (ΔH_decomp) for the decomposition of sodium hydrogen carbonate (NaHCO₃) using Hess's Law, we need to consider the reactions and their enthalpies.

The decomposition of sodium hydrogen carbonate can be represented as:

\[ 2 \text{NaHCO}_3 (s) \rightarrow \text{Na}_2\text{CO}_3 (s) + \text{CO}_2 (g) + \text{H}_2O (g) \]

We can find ΔH_decomp using the following reactions:

1. The first reaction: \[ \text{NaHCO}_3 (s) + \text{HCl} (aq) \rightarrow \text{NaCl} (aq) + \text{CO}_2 (g) + \text{H}_2O (g) \] Given: \( \Delta H_1 = +15.85 , \text{kJ/mol} \)

2. The second reaction: \[ \text{Na}_2\text{CO}_3 (s) + 2 , \text{HCl} (aq) \rightarrow 2 , \text{NaCl} (aq) + \text{CO}_2 (g) + \text{H}_2O (g) \] Given: \( \Delta H_2 = +13.50 , \text{kJ/mol} \)

Using Hess's Law:

Hess's Law states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.

Now, we'll manipulate the two given reactions to find the decomposition enthalpy of sodium hydrogen carbonate.

Starting from the second reaction, we need to reverse it because we want to go from sodium carbonate back to sodium hydrogen carbonate:

\[ 2 , \text{NaCl} (aq) + \text{CO}_2 (g) + \text{H}_2O (g) \rightarrow \text{Na}_2\text{CO}_3 (s) + 2 , \text{HCl} (aq) \] The enthalpy change for this reversed reaction will have the opposite sign: \[ \Delta H = -13.50 , \text{kJ/mol} , \text{(for 2 moles)} \]

For 1 mole of \(\text{Na}_2\text{CO}_3\): \[ \Delta H = -\frac{13.50}{2} , \text{kJ/mol} = -6.75 , \text{kJ/mol} \]

Now, we add the first reaction:

\[ \text{NaHCO}_3 (s) + \text{HCl} (aq) \rightarrow \text{NaCl} (aq) + \text{CO}_2 (g) + \text{H}_2O (g) \quad \Delta H = +15.85 , \text{kJ/mol} \]

When we add these two reactions together, the \(\text{NaCl} (aq)\), \(\text{CO}_2 (g)\), and \(\text{H}_2O (g)\) will cancel out.

Thus, we get:

\[ \text{NaHCO}_3 (s) + \text{HCl} (aq) + 2 , \text{NaCl} (aq) + \text{CO}_2 (g) + \text{H}_2O (g) \rightarrow \text{Na}_2\text{CO}_3 (s) + 2 , \text{HCl} (aq) \]

Summing Up the Enthalpy Changes:

The total ΔH (from the decomposition) will be:

\[ \Delta H_{decomp} = -6.75 , \text{kJ/mol} + 15.85 , \text{kJ/mol} \]

Calculating this:

\[ \Delta H_{decomp} = 9.10 , \text{kJ/mol} \]

However, since this represents the enthalpy change for the decomposition of 2 moles of NaHCO₃, we need to divide by 2 to find the per mole change:

\[ \Delta H_{decomp} = \frac{9.10 , \text{kJ/mol}}{2} = 4.55 , \text{kJ/mol} \]

Final Result:

The theoretical molar enthalpy change (ΔH_decomp) for the decomposition of sodium hydrogen carbonate is:

\[ \Delta H_{decomp} = 4.55 , \text{kJ/mol} \]