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Standard enthalpy of formation is when one mole of compound is formed from its constituent elements.

Example : $H_2$(g) + $O_2$(g) —>$ H_2O$(L).

But we don’t say the same for $CaO + CO_2$ —> $CaCO_3$.

It is because $CaCO_3$ is not formed it’s constituent elements but from other compounds.

Q1: From other compounds , does it mean the most simplest state I.e it is not formed from Ca + $O_2$.

I want to confirm if that’s all the meaning of standard enthalpy of formation. I am getting a lot confused over it.

Q2 Also , for the equation above. We can never find standard enthalpy of formation but can find the standard enthalpy of reaction I.e addition of $\delta$H of $CaO $ in solid state + $CO_2$ in gaseous state ?

Edit: enter image description here

S.M.T
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1 Answers1

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I prefer this definition for Hf: "The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states."

Simply put though heat of formation is a tool we use to quickly calculate heat of reaction for a compound. If I know the heat of formation for each reactant and product I can use the formula below (Hess's Law) to calculate heat of reaction.

heat of reaction

Once the calculation is done we will know if the chemical reaction is going to be endothermic or exothermic and to what magnitude.

For your calcium carbonate example you wouldn't need to tell me the heat of reaction I could simply look up the heat of formation for $CaO, CO_2, CaCO_3$ and do the math in the formula above.

$CaO: Ca(s) + (1/2) O_2(g) --> CaO (s) Hf = -635.5 $KJ/mol

$CO_2: C(s) + O_2(g) --> CO_2(g) Hf = -393.5 KJ/mol$

$CaCO_3:Ca(s)+C(s)+(3/2)O_2→CaCO_3(s) Hf = -1207.0 KJ/mol$

Given the reaction:

$CaO(s) + CO_2(g) --> CaCO_3(s)$

$Heat of reaction = (-1207.0)-(-635.5-393.5) = -178 KJ/mol$

Feynman137
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