Question

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

a) Calculate the molar enthalpy of decomposition of NaHCO₃ (ΔHdecomp). [6]

Answer 1

To calculate the molar enthalpy of decomposition of sodium hydrogen carbonate (NaHCO₃), we need to follow these steps:

1. **Calculate the number of moles of NaHCO₃ used**.
2. **Calculate the amount of heat absorbed by the water**.
3. **Calculate the molar enthalpy of decomposition from the heat and moles of NaHCO₃**.

### Step 1: Calculate the number of moles of NaHCO₃.

The molar mass of NaHCO₃ is calculated as follows:
- Na: 22.99 g/mol
- H: 1.01 g/mol
- C: 12.01 g/mol
- O: 16.00 g/mol × 3 = 48.00 g/mol

Calculating the total:
\[
\text{Molar mass of NaHCO}_3 = 22.99 + 1.01 + 12.01 + 48.00 = 84.01 \, \text{g/mol}
\]

Now, calculate the number of moles (\( n \)) of NaHCO₃:
\[
n = \frac{\text{mass}}{\text{molar mass}} = \frac{4.3 \, \text{g}}{84.01 \, \text{g/mol}} \approx 0.0511 \, \text{mol}
\]

### Step 2: Calculate the amount of heat absorbed by the water.

To find the heat absorbed by water (\( q \)), we use:
\[
q = m \cdot c \cdot \Delta T
\]
where:
- \( m \) = mass of water = volume × density = 20.0 mL × 1 g/mL = 20.0 g (assuming density of water is 1 g/mL)
- \( c \) = specific heat capacity of water = 4.18 J/g°C
- \( \Delta T \) = change in temperature = final temperature - initial temperature = 18.3°C - 29°C = -10.7°C (the temperature decreased, so it's negative)

Now plug in the values:
\[
q = 20.0 \, \text{g} \cdot 4.18 \, \text{J/g°C} \cdot (-10.7 \, \text{°C})
\]
\[
q = 20.0 \cdot 4.18 \cdot (-10.7) \approx -894.76 \, \text{J}
\]

(We take the absolute value since \( q \) represents the heat released from the reaction to the surroundings.)

### Step 3: Calculate the molar enthalpy of decomposition.

To find the molar enthalpy of decomposition (\( \Delta H_{\text{decomp}} \)), we need to divide the heat absorbed by the number of moles of NaHCO₃.

\[
\Delta H_{\text{decomp}} = \frac{q}{n}
\]
\[
\Delta H_{\text{decomp}} = \frac{-894.76 \, \text{J}}{0.0511 \, \text{mol}} \approx -17555.53 \, \text{J/mol}
\]

Converting to kJ/mol (since it's common to express ΔH in kJ):
\[
\Delta H_{\text{decomp}} \approx -17.56 \, \text{kJ/mol}
\]

### Final Answer:
The molar enthalpy of decomposition of NaHCO₃ is approximately **-17.56 kJ/mol**.