5. Reaction Enthalpy. and calculate AH for the following chemical equations. AH (kJ/mol): HCN = 130.5, Cas(PO4)2=-4120.8, H3PO4--1288, SiCl₁=-640.1, SiO₂= -910.9, Mg0=-601.6, HCl=-167.16 b) Ca,(PO₂)2(3) + H₂SO.(1) —— CaSO4(s) + H₂PO.(1)

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**5. Reaction Enthalpy.**

Calculate ΔH° for the following chemical equations. ΔH° (kJ/mol): HCN = 130.5, Ca₃(PO₄)₂ = −4120.8, H₃PO₄ = −1288, SiCl₄ = −640.1, SiO₂ = −910.9, MgO = −601.6, HCl = −167.16

**b)** \[ \square \, \text{Ca}_3(\text{PO}_4)_2(s) + \square \, \text{H}_2\text{SO}_4(l) \rightarrow \square \, \text{CaSO}_4(s) + \square \, \text{H}_3\text{PO}_4(l) \]

**Diagram Explanation:**
This is a chemical equation representing the reaction between calcium phosphate \(\text{Ca}_3(\text{PO}_4)_2\) and sulfuric acid \(\text{H}_2\text{SO}_4\) to form calcium sulfate \(\text{CaSO}_4\) and phosphoric acid \(\text{H}_3\text{PO}_4\). The squares indicate that you need to balance the equation by determining the correct stoichiometric coefficients for each chemical species.
Transcribed Image Text:**5. Reaction Enthalpy.** Calculate ΔH° for the following chemical equations. ΔH° (kJ/mol): HCN = 130.5, Ca₃(PO₄)₂ = −4120.8, H₃PO₄ = −1288, SiCl₄ = −640.1, SiO₂ = −910.9, MgO = −601.6, HCl = −167.16 **b)** \[ \square \, \text{Ca}_3(\text{PO}_4)_2(s) + \square \, \text{H}_2\text{SO}_4(l) \rightarrow \square \, \text{CaSO}_4(s) + \square \, \text{H}_3\text{PO}_4(l) \] **Diagram Explanation:** This is a chemical equation representing the reaction between calcium phosphate \(\text{Ca}_3(\text{PO}_4)_2\) and sulfuric acid \(\text{H}_2\text{SO}_4\) to form calcium sulfate \(\text{CaSO}_4\) and phosphoric acid \(\text{H}_3\text{PO}_4\). The squares indicate that you need to balance the equation by determining the correct stoichiometric coefficients for each chemical species.
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Here we are required to find the Enthalpy change for the given double displacement reaction. 

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