Standard enthalpy change of a reaction ( H r °)= Sum of standard enthalpy change

Question

Standard enthalpy change of a reaction ( Hr°)= Sum of standard enthalpy change of formation of products (Hf°) -Sum of standard enthalpy change of formation of reactants (Hf°)

The values for Hf° can be found thermodynamic property databases

Hf° Mg(OH)2 (s) = -924.7 kJ/mole

Hf° HCl (g) = -92.3 kJ/mole

Hf° MgCl2(s) = -641.8 kJ/mole

Hf° H2O(g) = -241.8 kJ/mole

Here, for the reaction, Mg (OH)2(s) + 2 HCl(g) MgCl2(s) + 2 H2O(g)

Hreaction° = (-641.8 + 2 x -241.8) - (-924.7 + 2 x -92.3) = -1125.4 - (-1109.3) = -16 kJ

Explanation / Answer

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