The heat of formation of HCI at 348 K from the following data, will be: 0.5H2(g)+0.5Cl2(g)⟶HCl ΔH298∘=−22060cal .The mean heat capacities over this temperature range are :H2(g), Cp=6.82calmol−1K−1Cl2(g),Cp=7.71calmol−1K−1HCl(g),Cp=6.81calmol−1K−1

ΔCP( Reaction )=6.81−0.5(6.82+7.71)=−0.455The given Formula is :ΔH348∘=ΔH298∘+ΔCPT2−T1=−22060−0.455×50=−22083cal

The heat of formation of HCI at 348 K from the following data, will be: 0.5H2(g)+0.5Cl2(g)⟶HCl ΔH298∘=−22060cal .The mean heat capacities over this temperature range are :H2(g), Cp=6.82calmol−1K−1Cl2(g),Cp=7.71calmol−1K−1HCl(g),Cp=6.81calmol−1K−1

ΔCP( Reaction )=6.81−0.5(6.82+7.71)=−0.455The given Formula is :ΔH348∘=ΔH298∘+ΔCPT2−T1=−22060−0.455×50=−22083cal

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The heat of formation of $HCI$ at $348 K$ from the following data, will be: $0.5 H_{2} (g) + 0.5 {Cl}_{2} (g) ⟶ HCl Δ H_{298}^{\circ} = - 22060 cal$ .The mean heat capacities over this temperature range are :
$H_{2} (g), C_{p} = 6.82 cal {mol}^{- 1} K^{- 1}$
${Cl}_{2} (g), C_{p} = 7.71 cal {mol}^{- 1} K^{- 1}$
$HCl (g), C_{p} = 6.81 cal {mol}^{- 1} K^{- 1}$

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$- 20095 cal$

$- 32758 cal$

$- 37725 cal$

$- 22083 cal$

answer is D.

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Detailed Solution

$Δ C_{P} (Reaction) = 6.81 - 0.5 (6.82 + 7.71) = - 0.455$

The given Formula is :

$Δ H_{348}^{\circ} = Δ H_{298}^{\circ} + Δ C_{P} (T_{2} - T_{1})$

$= - 22060 - 0.455 \times 50$

$= - 22083 cal$

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The heat of formation of HCI at 348 K from the following data, will be: 0.5H2(g)+0.5Cl2(g)⟶HCl ΔH298∘=−22060cal .The mean heat capacities over this temperature range are :H2(g), Cp=6.82calmol−1K−1Cl2(g),Cp=7.71calmol−1K−1HCl(g),Cp=6.81calmol−1K−1

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