Bond energies of H - H and CI - CI are 430 kJ mol-1 and 242 kJ mol-1 respectively- - Hf for HCl is 91 kJ mol-1 - What will be the bond energy of H - Cl bond per mole value-

The correct option is C 245 kJ 1/2 H_2 + 1/2 Cl_2 → HCl Δ H_f(HCl) = 91 kJ mol^ 1 , H H = 430 kJ mol^ 1 , Cl Cl = 242 kJ mol^ 1 9 ...

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Byju's Answer

Standard XII

Chemistry

Enthalpy

Bond energies...

Question

Bond energies of H - H and CI - CI are $430 k J m o l^{- 1}$ and $242 k J m o l^{- 1}$ respectively. $Δ H_{f}$ for HCl is $91 k J m o l^{- 1}$ . What will be the bond energy of H - Cl bond (per mole value)?

672 kJ

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182 kJ

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245 kJ

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88 kJ

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Solution

The correct option is C 245 kJ
$\frac{1}{2} H_{2} + \frac{1}{2} C l_{2} \to H C l$
$Δ H_{f} (H C l) = 91 k J m o l^{- 1}$ , H - H = 430 kJ $m o l^{- 1}$ , Cl - Cl = 242 kJ $m o l^{- 1}$
$91 = \frac{1}{2} \times 430 + \frac{1}{2} \times 242 - B . E . (H - C l)$
B.E. (H-Cl) = 245 kJ

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Bond energies of H - H and CI - CI are 430 kJmol−1 and 242 kJmol−1 respectively. ΔHf for HCl is 91 kJmol−1 . What will be the bond energy of H - Cl bond (per mole value)?

The correct option is C 245 kJ12H2+12Cl2→HClΔHf(HCl)=91 kJ mol−1, H - H = 430 kJ mol−1, Cl - Cl = 242 kJ mol−191=12×430+12×242−B.E.(H−Cl)B.E. (H-Cl) = 245 kJ

Bond energies of H - H and CI - CI are $430 k J m o l^{- 1}$ and $242 k J m o l^{- 1}$ respectively. $Δ H_{f}$ for HCl is $91 k J m o l^{- 1}$ . What will be the bond energy of H - Cl bond (per mole value)?

The correct option is C 245 kJ
$\frac{1}{2} H_{2} + \frac{1}{2} C l_{2} \to H C l$
$Δ H_{f} (H C l) = 91 k J m o l^{- 1}$ , H - H = 430 kJ $m o l^{- 1}$ , Cl - Cl = 242 kJ $m o l^{- 1}$
$91 = \frac{1}{2} \times 430 + \frac{1}{2} \times 242 - B . E . (H - C l)$
B.E. (H-Cl) = 245 kJ