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Calculate the standard reaction enthalpy for the reaction of calcite with hydrochloric acid

CaCO3(s) + 2 HCl(aq) →

CaCl2(aq) + H2O(ℓ) + CO2(g)

The standard enthalpies of formation are: 

for CaCl2(aq) : −877.1 kJ/mol; 

for H2O(ℓ) : −285.83 kJ/mol; 

for CO2(g) : −393.51 kJ/mol; 

for CaCO3(s) : −1206.9 kJ/mol; and 

for HCl(aq) : −167.16 kJ/mol.

1. −72.7 kJ/mol 

2. −38.2 kJ/mol 

3. −215 kJ/mol 

4. −116 kJ/mol 

5. −165 kJ/mol 

6. −15.2 kJ/mol 

 7. −98.8 kJ/mol

1 Answer

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Best answer

Correct option (6) −15.2 kJ/mol

Explanation:

We use Hess’ Law:

∆H◦ = ∑ n ∆H◦ j,prod − ∑ n ∆H◦ j,reac

= ∆H◦ f, CaCl2(aq) + ∆H◦ f, H2O(ℓ)

+ ∆H◦ f,CO2(g) − [ ∆H◦ f, CaCO3(s) 2 + (∆H◦ f, HCl(aq)]

= −877.1 kJ/mol + (−285.83 kJ/mol)

+ (−393.51 kJ/mol) − [ −1206.9 kJ/mol + 2 (−167.16 kJ/mol)]

= −15.22 kJ/mol .

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