1. Tardigrade
  2. Question
  3. Chemistry
  4. For the given reactions, H 2(g)+ F 2(g) arrow 2 HF (g) ; , Δ H°=-124 kcal H 2(g) arrow 2 H (g) ; Δ H°=104 kcal F2(g) arrow 2 F(g);, Δ H°=37.8 kcal then the value of Δ H° for H (g)+ F (g) arrow HF (g) is

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Q. For the given reactions,
$H _{2(g)}+ F _{2(g)} \rightarrow 2 HF _{(g)} ; ,\,\,\,\, \Delta H^{\circ}=-124\, kcal$
$H _{2(g)} \rightarrow 2 H _{(g)} ;\,\,\,\, \Delta H^{\circ}=104\, kcal$
$F_{2(g)} \rightarrow 2 F_{(g)};,\,\,\,\, \Delta H^{\circ}=37.8\, kcal$
then the value of $\Delta H^{\circ}$ for $H _{(g)}+ F _{(g)} \rightarrow HF _{(g)}$ is

Thermodynamics

Solution:

For the reaction,

$H _{2(g)}+ F _{2(g)} \rightarrow 2 HF _{(g)} ; ,\,\,\,\, \Delta H^{\circ}=-124\, kcal$

$\Delta H^{\circ}=\Sigma B . E .$ (reactants) $-\Sigma B . E .$ (products)

or $-124=\Delta H_{ H - H }+\Delta H_{ F - F }-2 \Delta H_{ H - F }$

$=104+37.8-2 \Delta H_{ H - F }$

$\therefore 2 \Delta H_{ H - F }^{\circ}=104+37.8+124=265.8\, kcal$

Bond energy of $H - F =\frac{265.8}{2}=132.9\, kcal$

$\therefore \Delta H^{\circ}$ for the given reaction $=-132.9 \,kcal$