Standard molar enthalpy of formation, $△_{f} H^{⊖}$ is just a special case of enthalpy of reaction , $△_{r} H^{⊖}$ . Is the $△_{r} H^{⊖}$ for the following reaction same as $△_{f} H^{⊖}$ ?Give reason for your answer.

$C a O (s) + C O (g) \to C a C O_{3} (s)$ ;

$△_{f} H^{⊖} = - 178.3 k j m o l^{- 1}$

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Solution

Standard enthalpy of formation is the enthalpy of the reaction when one mole of a compound is formed from its constituent elements being each in their respective standard state.

Here, $C a C O_{3}$ is not formed from its constituent element. Therefore, for the given reaction

$△_{r} H^{⊖} \neq △_{f} H^{⊖}$

For $△_{r} H^{⊖} = △_{f} H^{⊖}$ , the equation should be:

$C a (s) + C (s) + \frac{3}{2} O_{2} (g) \to C a C O_{3} (s)$ ;

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Standard molar enthalpy of formation, $Δ_{f} H^{⊖}$ is just a special case of enthalpy of reaction, $Δ_{r} H^{⊖}$ . Is the $Δ_{r} H^{⊖}$ for the following reaction same as $Δ_{f} H^{⊖}$ ? Give reason for your answer.
$C a 0 (s) + C O_{2} (g) \to C a C O_{3} (s); Δ_{f} H^{⊖} = - 178.3 k J m o l^{- 1}$