Question
. In the Solvay process for the production of sodium carbonate (or washing soda), one step is the
decomposition of sodium hydrogen carbonate.
2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(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 29oC to 18.3oC.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]
decomposition of sodium hydrogen carbonate.
2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(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 29oC to 18.3oC.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]
Answers
Answered by
GPT-4o mini
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 can manipulate the provided reactions and their associated enthalpy changes.
The decomposition reaction we want is:
\[
2 \text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{CO}_2(g) + \text{H}_2\text{O}(g)
\]
We have the following 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.
\[
\text{Na}_2\text{CO}_3(s) + 2 \text{HCl}(aq) \rightarrow 2 \text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(g) \quad \Delta H_2 = +13.50 \, \text{kJ/mol}
\]
To find the enthalpy change for the decomposition reaction (ΔH_decomp), we will need to combine these reactions in such a way that they yield the desired equation.
Here’s how we can manipulate the reactions:
1. Start with reaction 1 as it is, but we need two moles of NaHCO3 for our target equation, so we multiply it by 2:
\[
2 \text{NaHCO}_3(s) + 2 \text{HCl}(aq) \rightarrow 2 \text{NaCl}(aq) + 2 \text{CO}_2(g) + 2 \text{H}_2\text{O}(g) \quad \Delta H_1 = 2 \times 15.85 \, \text{kJ/mol} = +31.70 \, \text{kJ}
\]
2. Reverse reaction 2 to represent the formation of Na2CO3 from NaHCO3, which allows for the CO2 and H2O to produce:
\[
2 \text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(g) \rightarrow \text{Na}_2\text{CO}_3(s) + 2 \text{HCl}(aq) \quad \Delta H_2 = -13.50 \, \text{kJ}
\]
Now, if we add these two manipulated reactions together, we have:
1.
\[
2 \text{NaHCO}_3(s) + 2 \text{HCl}(aq) \rightarrow 2 \text{NaCl}(aq) + 2 \text{CO}_2(g) + 2 \text{H}_2\text{O}(g) \quad (+31.70 \, \text{kJ})
\]
2.
\[
2 \text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(g) \rightarrow \text{Na}_2\text{CO}_3(s) + 2 \text{HCl}(aq) \quad (-13.50 \, \text{kJ})
\]
Adding these two reactions cancels out the sodium chloride (NaCl) and HCl terms, giving:
\[
2 \text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{CO}_2(g) + \text{H}_2\text{O}(g)
\]
Now we can calculate ΔH_decomp:
\[
\Delta H_{\text{decomp}} = +31.70 \, \text{kJ} - 13.50 \, \text{kJ} = +18.20 \, \text{kJ}
\]
Finally, to find the molar enthalpy change for one mole of NaHCO3 decomposing, we divide the result by 2:
\[
\Delta H_{\text{decomp}} = \frac{+18.20 \, \text{kJ}}{2} = +9.10 \, \text{kJ/mol}
\]
Thus, the theoretical value of the molar enthalpy change (ΔH_decomp) for the decomposition of sodium hydrogen carbonate (NaHCO3) is **+9.10 kJ/mol**.
The decomposition reaction we want is:
\[
2 \text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{CO}_2(g) + \text{H}_2\text{O}(g)
\]
We have the following 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.
\[
\text{Na}_2\text{CO}_3(s) + 2 \text{HCl}(aq) \rightarrow 2 \text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(g) \quad \Delta H_2 = +13.50 \, \text{kJ/mol}
\]
To find the enthalpy change for the decomposition reaction (ΔH_decomp), we will need to combine these reactions in such a way that they yield the desired equation.
Here’s how we can manipulate the reactions:
1. Start with reaction 1 as it is, but we need two moles of NaHCO3 for our target equation, so we multiply it by 2:
\[
2 \text{NaHCO}_3(s) + 2 \text{HCl}(aq) \rightarrow 2 \text{NaCl}(aq) + 2 \text{CO}_2(g) + 2 \text{H}_2\text{O}(g) \quad \Delta H_1 = 2 \times 15.85 \, \text{kJ/mol} = +31.70 \, \text{kJ}
\]
2. Reverse reaction 2 to represent the formation of Na2CO3 from NaHCO3, which allows for the CO2 and H2O to produce:
\[
2 \text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(g) \rightarrow \text{Na}_2\text{CO}_3(s) + 2 \text{HCl}(aq) \quad \Delta H_2 = -13.50 \, \text{kJ}
\]
Now, if we add these two manipulated reactions together, we have:
1.
\[
2 \text{NaHCO}_3(s) + 2 \text{HCl}(aq) \rightarrow 2 \text{NaCl}(aq) + 2 \text{CO}_2(g) + 2 \text{H}_2\text{O}(g) \quad (+31.70 \, \text{kJ})
\]
2.
\[
2 \text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(g) \rightarrow \text{Na}_2\text{CO}_3(s) + 2 \text{HCl}(aq) \quad (-13.50 \, \text{kJ})
\]
Adding these two reactions cancels out the sodium chloride (NaCl) and HCl terms, giving:
\[
2 \text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{CO}_2(g) + \text{H}_2\text{O}(g)
\]
Now we can calculate ΔH_decomp:
\[
\Delta H_{\text{decomp}} = +31.70 \, \text{kJ} - 13.50 \, \text{kJ} = +18.20 \, \text{kJ}
\]
Finally, to find the molar enthalpy change for one mole of NaHCO3 decomposing, we divide the result by 2:
\[
\Delta H_{\text{decomp}} = \frac{+18.20 \, \text{kJ}}{2} = +9.10 \, \text{kJ/mol}
\]
Thus, the theoretical value of the molar enthalpy change (ΔH_decomp) for the decomposition of sodium hydrogen carbonate (NaHCO3) is **+9.10 kJ/mol**.