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The heat of reaction (in kcal) is minus: BaCl2(aq) + H2SO4(aq)  ⟶  BaSO4(s) + 2HCl(aq)  is__________.

Question

The heat of reaction (in kcal) is minus:

BaCl2(aq) + H2SO4(aq) ⟶ BaSO4(s) + 2HCl(aq)

is__________.

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Solution

To find the heat of reaction for the given chemical equation, we can use calorimetry or standard enthalpy changes if provided. However, since the specific value is not given in your question, I can guide you through the steps you would typically follow to determine it.

1. Break Down the Problem

You are provided with a reaction: BaCl2(aq)+H2SO4(aq)BaSO4(s)+2HCl(aq) \text{BaCl}_2(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{BaSO}_4(s) + 2\text{HCl}(aq) We need to determine the heat of reaction, ΔH \Delta H .

2. Relevant Concepts

The heat of reaction can typically be determined using standard enthalpy values of the reactants and products: ΔH=ΔHproductsΔHreactants \Delta H = \sum \Delta H_\text{products} - \sum \Delta H_\text{reactants} If the standard enthalpy of formation values (ΔHf \Delta H_f^\circ ) are known for each compound, we can apply this formula.

3. Analysis and Detail

  1. Identify the standard enthalpy of formation for each reactant and product:

    • ΔHf(BaCl2) \Delta H_f^\circ(\text{BaCl}_2)
    • ΔHf(H2SO4) \Delta H_f^\circ(\text{H}_2\text{SO}_4)
    • ΔHf(BaSO4) \Delta H_f^\circ(\text{BaSO}_4)
    • ΔHf(HCl) \Delta H_f^\circ(\text{HCl})
  2. Substitute these values into the equation for ΔH \Delta H .

Assuming the following values (for example purposes, you would need to check reference data for accurate values):

  • ΔHf(BaCl2) \Delta H_f^\circ(\text{BaCl}_2) = -100.0 kcal/mol
  • ΔHf(H2SO4) \Delta H_f^\circ(\text{H}_2\text{SO}_4) = -200.0 kcal/mol
  • ΔHf(BaSO4) \Delta H_f^\circ(\text{BaSO}_4) = -250.0 kcal/mol
  • ΔHf(HCl) \Delta H_f^\circ(\text{HCl}) = -30.0 kcal/mol

Now calculate: ΔHproducts=ΔHf(BaSO4)+2ΔHf(HCl)=250.0+2(30.0)=250.060.0=310.0 kcal \sum \Delta H_\text{products} = \Delta H_f^\circ(\text{BaSO}_4) + 2 \cdot \Delta H_f^\circ(\text{HCl}) = -250.0 + 2(-30.0) = -250.0 - 60.0 = -310.0 \text{ kcal} ΔHreactants=ΔHf(BaCl2)+ΔHf(H2SO4)=100.0200.0=300.0 kcal \sum \Delta H_\text{reactants} = \Delta H_f^\circ(\text{BaCl}_2) + \Delta H_f^\circ(\text{H}_2\text{SO}_4) = -100.0 - 200.0 = -300.0 \text{ kcal} Finally, applying these to the enthalpy equation: ΔH=310.0(300.0)=10.0 kcal \Delta H = -310.0 - (-300.0) = -10.0 \text{ kcal}

4. Verify and Summarize

  • Make sure each value used from the reference is accurate.
  • Confirm that the signs of enthalpy values are consistent with formation definitions.

Final Answer

The heat of reaction is approximately 10.0kcal-10.0 \, \text{kcal}.

Please verify with actual standard enthalpy data as this is a general guide using assumed values.

This problem has been solved

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