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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

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  • $\begingroup$ Which book has the examples you show? Are they from the same page, passage or chapter? Perhaps they are focusing on different things. $\endgroup$
    – Solar Mike
    Jun 13 at 15:13
  • $\begingroup$ I’ll paste them on the Q. $\endgroup$
    – S.M.T
    Jun 13 at 15:16
  • $\begingroup$ So is your first example incorrect ? Either missing 0.5 as the result is incorrect as you are missing some oxygen. $\endgroup$
    – Solar Mike
    Jun 13 at 17:45
  • $\begingroup$ @SolarMike Firs is correct. I’m asking for next one. That’s why I wrote but $\endgroup$
    – S.M.T
    Jun 13 at 17:53
  • $\begingroup$ It is incorrect, you have H2 + O2 to give H20 when the correct result would be H2O2 or even H2O + O. The4 second does balance but the first not. $\endgroup$
    – Solar Mike
    Jun 13 at 18:13
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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:

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

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

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  • $\begingroup$ Ok. So , if Q says to find $\delta$H_f of the above reaction. Then , should it be -1207 or -178 ? Also , For formation of CO2. delta Hf = delta Hr ? $\endgroup$
    – S.M.T
    Jun 14 at 10:45
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    $\begingroup$ @SrijanM.T For question one: The heat of formation of CaCO3 is -1207 KJ/mol. This is much more exothermic than the heat of reaction of CaCO3 from CO2 and CaO. For question two: the heat of reaction is equal to the heat of formation as you have stated. $\endgroup$
    – Feynman137
    Jun 14 at 12:49

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