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- Ca 0 s - CO 2 g - CaCO 3 s - - f H -178-3 kJ mol -1

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Standard XII

Chemistry

Heat of Formation

Standard mola...

Question

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.
$C a 0 (s) + C O_{2} (g) \to C a C O_{3} (s); Δ_{f} H^{⊖} = - 178.3 k J m o l^{- 1}$

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Solution

No, the $Δ_{r} H^{⊖}$ for the given reaction is not same as $Δ_{r} H^{⊖}$ . The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states (reference states) is called standard molar enthalpy of formation, $Δ_{r} H^{⊖}$ .
$C a (S) + C (s) + \frac{3}{2} O_{2} (g) \to C a C O_{3} (s); Δ_{f} H^{s}$
This reaction is different from the given reaction.
Hence, $Δ_{r} H^{\circ} \neq Δ_{f} H^{\circ}$

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Similar questions

Q. Follow data is given for the reaction:

C a C O_{3} ⇌ C a O (s) + C O_{2} (g)

Δ_{f} H^{⊖} [C a O (s)] = - 635.1 k J m o l^{- 1}

Δ_{f} H^{⊖} [C O_{2} (g)] = - 393.5 k J m o l^{- 1}

Δ_{f} H^{⊖} [C a C O_{3} (s)] = - 1206.9 k J m o l^{- 1}

Predict the effect of temperature on the equilibrium constant of the above reaction.

Q. Following data is given for the reaction:

C a C O_{3} (s) \to C a O (s) + C O_{2} (g)

Δ_{f} H^{⊝} [C a O (s)] = - 635.1 k J

m o l^{- 1}

Δ_{f} H^{⊝} [C O_{2} (g)] = - 393.5 k J

m o l^{- 1}

Δ_{f} H^{⊝} [C a C O_{3} (s)] = - 1206.9 k J

m o l^{- 1}

Predict the effect of temperature on the equilibrium constant of the above reaction.

Q. The

Δ_{f} H^{⊖}

for

C O_{2} (g), C O (g) a n d H_{2} O (g)

are

- 393.5, - 110.5 a n d - 214.8 k J m o l^{- 1}

, respectively. The standard enthalpy change

(i n k J m o l^{- 1})

for the reaction

C O_{2} (g) + H_{2} (g) ⟶ C O (g) + H_{2} O (g)

is:

Q. The

Δ_{f} H^{\circ}

for

C O_{2} (g), C O (g) a n d H_{2} O (g)

are

- 393.5, - 110.5

and

- 241.8 k J m o l^{- 1}

respectively. The standard enthalpy change (in kJ) for the reaction is:

C O_{2} (g) + H_{2} (g) ⟶ C O (g) + H_{2} O (g)

The value of $Δ_{f} H^{⊖}$ for $N H_{3}$ is -91.8kJ $m o l^{- 1}$ . Calculate enthalpy change for the following reaction.
$2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$

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Standard molar enthalpy of formation, ΔfH⊖ is just a special case of enthalpy of reaction, ΔrH⊖ . Is the ΔrH⊖ for the following reaction same as ΔfH⊖ ? Give reason for your answer. Ca0(s)+CO2(g)→CaCO3(s);ΔfH⊖=−178.3 kJ mol−1