Question
If this reaction was too difficult to conduct in a calorimeter, the following known reactions would be used
to determine ΔHdecomp
:
1. NaHCO3(s) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(g) ΔH1 = +15.85 kJ/mol
2. Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(g) ΔH2 = +13.50 kJ/mol
Use Hess’s law to determine the theoretical value of the molar enthalpy change (ΔHdecomp) for NaHCO3
. [4]
to determine ΔHdecomp
:
1. NaHCO3(s) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(g) ΔH1 = +15.85 kJ/mol
2. Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(g) ΔH2 = +13.50 kJ/mol
Use Hess’s law to determine the theoretical value of the molar enthalpy change (ΔHdecomp) for NaHCO3
. [4]
Answers
Answered by
GPT-4o mini
To determine the molar enthalpy change (ΔH_decomp) for the decomposition of NaHCO₃ using Hess’s law, we need to set up the reactions such that they combine to yield the desired decomposition reaction. The decomposition of sodium bicarbonate (NaHCO₃) can be represented as follows:
\[
\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{CO}_2(g) + \text{H}_2\text{O}(g) \quad (\Delta H_{\text{decomp}})
\]
We can use the provided reactions in reverse or straight form as needed.
Given reactions:
1. \(\text{NaHCO}_3(s) + \text{HCl}(aq) \rightarrow \text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(g) \quad (\Delta H_1 = +15.85 \, \text{kJ/mol})\)
2. \(\text{Na}_2\text{CO}_3(s) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(g) \quad (\Delta H_2 = +13.50 \, \text{kJ/mol})\)
### Step 1: Manipulate the reactions
1. We will **reverse reaction 1** because we want to start with NaHCO₃ instead of producing it:
\[
\text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(g) \rightarrow \text{NaHCO}_3(s) + \text{HCl}(aq) \quad (\Delta H = -15.85 \, \text{kJ/mol})
\]
2. We will **leave reaction 2 as is**, but we will only need half of it because we have only one mole of NaHCO₃ in the decomposition:
\[
\text{Na}_2\text{CO}_3(s) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(g) \quad (\Delta H_2 = +13.50 \, \text{kJ/mol})
\]
To get half of this reaction, we will divide everything by 2:
\[
\frac{1}{2}\text{Na}_2\text{CO}_3(s) + \text{HCl}(aq) \rightarrow \text{NaCl}(aq) + \frac{1}{2}\text{CO}_2(g) + \frac{1}{2}\text{H}_2\text{O}(g) \quad (\Delta H = \frac{13.50}{2} = +6.75 \, \text{kJ/mol})
\]
### Step 2: Combine the reactions
Now we can combine the two reactions. The combination will have:
- **Left Side**: \(\text{NaHCO}_3 + \text{Na}_2\text{CO}_3 + 2\text{HCl}\)
- **Right Side**: 2\(\text{NaCl} + \text{CO}_2 + \text{H}_2\text{O}\)
Writing the overall balanced equation for our interest:
\[
\text{NaHCO}_3(s) \rightarrow \frac{1}{2} \text{Na}_2\text{CO}_3(s) + \text{CO}_2(g) + \text{H}_2\text{O}(g)
\]
### Step 3: Calculate ΔH_decomp
Now we can add the enthalpy changes:
\[
\Delta H_{\text{decomp}} = (-15.85 \, \text{kJ/mol}) + (+6.75 \, \text{kJ/mol}) = -15.85 + 6.75 = -9.10 \, \text{kJ/mol}
\]
Thus, the theoretical value of the molar enthalpy change for the decomposition of NaHCO₃ is:
\[
\boxed{-9.10 \, \text{kJ/mol}}
\]
\[
\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{CO}_2(g) + \text{H}_2\text{O}(g) \quad (\Delta H_{\text{decomp}})
\]
We can use the provided reactions in reverse or straight form as needed.
Given reactions:
1. \(\text{NaHCO}_3(s) + \text{HCl}(aq) \rightarrow \text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(g) \quad (\Delta H_1 = +15.85 \, \text{kJ/mol})\)
2. \(\text{Na}_2\text{CO}_3(s) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(g) \quad (\Delta H_2 = +13.50 \, \text{kJ/mol})\)
### Step 1: Manipulate the reactions
1. We will **reverse reaction 1** because we want to start with NaHCO₃ instead of producing it:
\[
\text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(g) \rightarrow \text{NaHCO}_3(s) + \text{HCl}(aq) \quad (\Delta H = -15.85 \, \text{kJ/mol})
\]
2. We will **leave reaction 2 as is**, but we will only need half of it because we have only one mole of NaHCO₃ in the decomposition:
\[
\text{Na}_2\text{CO}_3(s) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(g) \quad (\Delta H_2 = +13.50 \, \text{kJ/mol})
\]
To get half of this reaction, we will divide everything by 2:
\[
\frac{1}{2}\text{Na}_2\text{CO}_3(s) + \text{HCl}(aq) \rightarrow \text{NaCl}(aq) + \frac{1}{2}\text{CO}_2(g) + \frac{1}{2}\text{H}_2\text{O}(g) \quad (\Delta H = \frac{13.50}{2} = +6.75 \, \text{kJ/mol})
\]
### Step 2: Combine the reactions
Now we can combine the two reactions. The combination will have:
- **Left Side**: \(\text{NaHCO}_3 + \text{Na}_2\text{CO}_3 + 2\text{HCl}\)
- **Right Side**: 2\(\text{NaCl} + \text{CO}_2 + \text{H}_2\text{O}\)
Writing the overall balanced equation for our interest:
\[
\text{NaHCO}_3(s) \rightarrow \frac{1}{2} \text{Na}_2\text{CO}_3(s) + \text{CO}_2(g) + \text{H}_2\text{O}(g)
\]
### Step 3: Calculate ΔH_decomp
Now we can add the enthalpy changes:
\[
\Delta H_{\text{decomp}} = (-15.85 \, \text{kJ/mol}) + (+6.75 \, \text{kJ/mol}) = -15.85 + 6.75 = -9.10 \, \text{kJ/mol}
\]
Thus, the theoretical value of the molar enthalpy change for the decomposition of NaHCO₃ is:
\[
\boxed{-9.10 \, \text{kJ/mol}}
\]