Calculate Delta H^{circ} for the reaction,Na_ | vedclass

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(B) We need to find $\Delta H^{\circ}$ for the reaction: $Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$.Given equations:$(A) \ Na_{(s)} + H_2

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$-581 \ kJ$

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(B) We need to find $\Delta H^{\circ}$ for the reaction: $Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$.Given equations:$(A) \ Na_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(s)} + \frac{1}{2} H_{2(g)} \quad \Delta H^{\circ} = -146 \ kJ$$(B) \ Na_2SO_{4(s)} + H_2O_{(l)} \longrightarrow 2NaOH_{(s)} + SO_{3(g)} \quad \Delta H^{\circ} = +418 \ kJ$$(C) \ 2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H^{\circ} = +259 \ kJ$To get the target reaction,perform the operation: $2 	imes (A) + \frac{1}{2} 	imes (C) - (B)$.$2 	imes [Na_{(s)} + H_2O_{(l)}$ $\longrightarrow NaOH_{(s)} + \frac{1}{2} H_{2(g)}] \implies 2Na_{(s)} + 2H_2O_{(l)}$ $\longrightarrow 2NaOH_{(s)} + H_{2(g)} \quad \Delta H^{\circ} = 2 	imes (-146) = -292 \ kJ$$\frac{1}{2} 	imes [2Na_2O_{(s)} + 2H_{2(g)}$ $\longrightarrow 4Na_{(s)} + 2H_2O_{(l)}] \implies Na_2O_{(s)} + H_{2(g)}$ $\longrightarrow 2Na_{(s)} + H_2O_{(l)} \quad \Delta H^{\circ} = \frac{259}{2} = 129.5 \ kJ$$-(B) \implies 2NaOH_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)} + H_2O_{(l)} \quad \Delta H^{\circ} = -418 \ kJ$Adding these: $(2Na_{(s)} + 2H_2O_{(l)} + Na_2O_{(s)} + H_{2(g)} + 2NaOH_{(s)} + SO_{3(g)})$ $\longrightarrow (2NaOH_{(s)} + H_{2(g)} + 2Na_{(s)} + H_2O_{(l)} + Na_2SO_{4(s)} + H_2O_{(l)})$Simplifying gives: $Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$.$\Delta H^{\circ} = -292 + 129.5 - 418 = -580.5 \ kJ \approx -581 \ kJ$.

Molecule	$\Delta_f H^0 (kJ \ mol^{-1})$
$C_2H_5OH_{(l)}$	$-280$
$CO_{2(g)}$	$-400$
$H_2O_{(l)}$	$-290$

Similar Questions

What is the enthalpy change for $2H_2O_{2(l)} \to 2H_2O_{(l)} + O_{2(g)}$ if the heats of formation of $H_2O_{2(l)}$ and $H_2O_{(l)}$ are $-188 \ kJ/mol$ and $-286 \ kJ/mol$ respectively?

Calculate $\Delta_r H$ $(kJ \ mol^{-1})$ of the following reaction:
$C_2H_5OH_{(l)} + \frac{7}{2}O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$
Molecule $\Delta_f H^0 (kJ \ mol^{-1})$
$C_2H_5OH_{(l)}$ $-280$
$CO_{2(g)}$ $-400$
$H_2O_{(l)}$ $-290$

The bond energies of $H-H$ and $Cl-Cl$ are $430 \ kJ \ mol^{-1}$ and $240 \ kJ \ mol^{-1}$ respectively. If the $\Delta H_f$ (enthalpy of formation) of $HCl$ is $-90 \ kJ \ mol^{-1}$,what is the $H-Cl$ bond energy in $kJ \ mol^{-1}$?

Calculate the bond enthalpy of the $H-Cl$ bond from the following reaction:
$H_{2(g)} + Cl_{2(g)} \rightarrow 2 HCl_{(g)}$,$\Delta_{r} H^{\circ} = -185 \ kJ \ mol^{-1}$
(Given bond enthalpies of $H-H$ and $Cl-Cl$ bonds are $435.0 \ kJ \ mol^{-1}$ and $244 \ kJ \ mol^{-1}$ respectively.)

Enthalpy of formation is a special case of enthalpy of reaction. Which of the following reactions does $NOT$ represent the enthalpy of formation of the product?

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