The enthalpy change of which reaction corresp | vedclass

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Question

(A) The standard enthalpy of formation $(\Delta H_f^o)$ is defined as the enthalpy change when $1 \ mol$ of a compound is formed from its constituent

Vedclass · Accepted Answer

$2Na_{(s)} + C_{(s)} + \frac{3}{2} O_{2(g)} 	o Na_2CO_{3(s)}$

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(A) The standard enthalpy of formation $(\Delta H_f^o)$ is defined as the enthalpy change when $1 \ mol$ of a compound is formed from its constituent elements in their most stable standard states at $298 \ K$ and $1 \ bar$ pressure.For $Na_2CO_{3(s)}$,the constituent elements are sodium $(Na)$,carbon $(C)$,and oxygen $(O_2)$.The standard states are $Na_{(s)}$,$C_{(graphite)}$,and $O_{2(g)}$.Therefore,the reaction is: $2Na_{(s)} + C_{(s)} + \frac{3}{2} O_{2(g)} 	o Na_2CO_{3(s)}$.

Bond	Bond energy $(kJ \ mol^{-1})$
$C-H$	$413$
$C-C$	$348$
$C=C$	$610$
$H-H$	$436$

The enthalpy change of which reaction corresponds to $\Delta H_f^o$ for $Na_2CO_{3(s)}$ at $298 \ K$?

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With the help of the following data,find out the change in heat content for the reaction:
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Bond Bond energy $(kJ \ mol^{-1})$

$C-H$ $413$

$C-C$ $348$

$C=C$ $610$

$H-H$ $436$

Given that the bond energy of $H-H$ bonds is $436 \ kJ/mol$,$O-H$ bonds is $464 \ kJ/mol$,and $O=O$ bonds is $496 \ kJ/mol$,what is the approximate heat of reaction for $2H_2 + O_2 \longrightarrow 2H_2O$ in $kJ/mol$?

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