Given- HClg - Hg - Clg at 290 K Given- rHo298 - 432-952 kJ also- Ho298 - Ho0-kJ mol-1 - Ho290 -kJ mol-1where Ho0-kJ mol-1 is the enthalpy at 0 KGiven- - HCl H Cl- 8-544 5-194 6-145 -Thus- for the given reaction rHo0 at 0 K is-

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Standard XII

Chemistry

Heat of Reaction

Given, HClg →...

Question

Given, $H C l (g) \to H (g) + C l (g)$ at 290 K
Given, $△_{r} H_{298}^{o} = 432.952 k J$ also, $(H_{298}^{o} - H_{0}^{o}) / k J m o l^{- 1} = H_{290}^{o} / k J m o l^{- 1}$
where $H_{0}^{o} / k J m o l^{- 1}$ is the enthalpy at 0 K
Given,
$\begin{matrix} H C l & H & C l 8.544 & 5.194 & 6.145 \end{matrix}$
Thus, for the given reaction $△_{r} H_{0}^{o}$ (at 0 K) is:

428.136

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430.157

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431.961

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- 431.961

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Solution

The correct option is B $430.157$ kJ
Given, $△ H = (H_{298}^{o} - H_{0}^{o})$
$[H (g) + C l (g) - H C l (g)]$
Enthalpy of formation is given as $Δ H_{p r o d u c t} - Δ H_{r e a c t a n t}$
$= (5.194 + 6.145) - 8.544$
$△ H_{298}^{o} - △ H_{0}^{o} = 2.795 k J$
$∴ △ H_{0}^{o} = △ H_{298}^{o} - 2.795$
$= 432.952 - 2.795 = 430.157 k J$

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Similar questions

Q. Given,

H C l (g) \to H (g) + C l (g)

at 290 K
Given,

△_{r} H_{298}^{o} = 432.952 k J

also,

(H_{298}^{o} - H_{0}^{o}) / k J m o l^{- 1} = H_{290}^{o} / k J m o l^{- 1}

where

H_{0}^{o} / k J m o l^{- 1}

is the enthalpy at 0 K
Given,

\begin{matrix} H C l & H & C l 8.544 & 5.194 & 6.145 \end{matrix}

Thus, for the given reaction

△_{r} H_{0}^{o}

(at 0 K) is:

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and

△ S^{o} = - 198.3 J K^{- 1} m o l^{- 1}

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Heat of Reaction

Standard XII Chemistry

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Given, HCl(g)→H(g)+Cl(g) at 290 K Given, △rHo298=432.952 kJ also, (Ho298−Ho0)/kJ mol−1=Ho290/kJ mol−1 where Ho0/kJ mol−1 is the enthalpy at 0 K Given, HClHCl8.5445.1946.145 Thus, for the given reaction △rHo0 (at 0 K) is:

The correct option is B 430.157 kJGiven, △H=(Ho298−Ho0) [H(g)+Cl(g)−HCl(g)] Enthalpy of formation is given as ΔHproduct−ΔHreactant =(5.194+6.145)−8.544 △Ho298−△Ho0=2.795 kJ ∴△Ho0=△Ho298−2.795 =432.952−2.795=430.157 kJ