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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]

1 answer

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:

\[ \boxed{-9.10 , \text{kJ/mol}} \]