Given below are two statements:Assertion (A): | vedclass

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(A) The enthalpy of formation of a substance is defined as the change in enthalpy when $1 \ mol$ of a substance is formed from its constituent element

Vedclass · Accepted Answer

Both $(A)$ and $(R)$ are correct and $(R)$ is the correct explanation of $(A)$

Vedclass · Answer

(A) The enthalpy of formation of a substance is defined as the change in enthalpy when $1 \ mol$ of a substance is formed from its constituent elements in their most stable states at standard conditions ($298 \ K$ and $1 \ bar$ pressure).Since graphite is the most stable allotrope of carbon at standard conditions,its standard enthalpy of formation is defined as zero.Therefore,the Assertion $(A)$ is correct because it states the standard enthalpy of formation of graphite is zero.The Reason $(R)$ is also correct because the stability of graphite as the most stable allotrope is the fundamental reason why its enthalpy of formation is taken as zero.Thus,$(R)$ is the correct explanation of $(A)$.

Given below are two statements:
Assertion $(A)$: The enthalpy of formation of graphite is taken as zero.
Reason $(R)$: Graphite is the thermodynamically most stable allotrope of carbon.
The correct answer is:

Similar Questions

The standard enthalpy of formation of $NH_3$ is $-46.0 \, kJ \, mol^{-1}$. If the enthalpy of atomization of $H_2$ is $436 \, kJ \, mol^{-1}$ and that of $N_2$ is $712 \, kJ \, mol^{-1}$,what is the average bond enthalpy of the $N-H$ bond in $NH_3$ in $kJ \, mol^{-1}$?

Based on the bond enthalpy $(B.E.)$ values given,the standard enthalpy of formation $(\Delta_fH^o)$ of $N_2H_{4(g)}$ is ...... $kJ\ mol^{-1}$.
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The enthalpy of combustion of benzene from the following data will be
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.....$kJ$

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Given below are two statements:Assertion $(A)$: The enthalpy of formation of graphite is taken as zero.Reason $(R)$: Graphite is the thermodynamically most stable allotrope of carbon.The correct answer is: