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Enthalpy

Bond energies...

Question

Bond energies of some bonds are given below:
Cl-Cl = 242.8 kJ $m o l^{- 1}$ , H-Cl = 431.8 kJ $m o l^{- 1}$ ,
O-H = 464 kJ $m o l^{- 1}$ , O=O = 442 kJ $m o l^{- 1}$
Using the B.E.s given, calculate $Δ H$ for the given reaction: $2 C l_{2} + 2 H_{2} O \to 4 H C l + O_{2}$

906 kJ

m o l^{- 1}

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172.4 kJ

m o l^{- 1}

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198.8 kJ

m o l^{- 1}

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442 kJ

m o l^{- 1}

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Solution

The correct option is C 172.4 kJ $m o l^{- 1}$
$2 C I_{2} + 2 H_{2} O \to 4 H C I + O_{2}$
From Hess' Law, $Δ H$ = B.E. of (2 X CI - CI) + (2 X 2 X O - H) - (4 X H - CI) + (O = O)
= 2 X 242.8 + 4 X 464 - 4 X 431.8 - 442
= 172.4 kJ $m o l^{- 1}$

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Bond energies of some bonds are given below:Cl-Cl = 242.8 kJ mol−1, H-Cl = 431.8 kJ mol−1,O-H = 464 kJ mol−1, O=O = 442 kJ mol−1Using the B.E.s given, calculate ΔH for the given reaction: 2Cl2+2H2O→4HCl+O2

The correct option is C 172.4 kJ mol−12CI2+2H2O→4HCI+O2 From Hess' Law, ΔH = B.E. of (2 X CI - CI) + (2 X 2 X O - H) - (4 X H - CI) + (O = O) = 2 X 242.8 + 4 X 464 - 4 X 431.8 - 442 = 172.4 kJ mol−1

Bond energies of some bonds are given below:
Cl-Cl = 242.8 kJ $m o l^{- 1}$ , H-Cl = 431.8 kJ $m o l^{- 1}$ ,
O-H = 464 kJ $m o l^{- 1}$ , O=O = 442 kJ $m o l^{- 1}$
Using the B.E.s given, calculate $Δ H$ for the given reaction: $2 C l_{2} + 2 H_{2} O \to 4 H C l + O_{2}$

The correct option is C 172.4 kJ $m o l^{- 1}$
$2 C I_{2} + 2 H_{2} O \to 4 H C I + O_{2}$
From Hess' Law, $Δ H$ = B.E. of (2 X CI - CI) + (2 X 2 X O - H) - (4 X H - CI) + (O = O)
= 2 X 242.8 + 4 X 464 - 4 X 431.8 - 442
= 172.4 kJ $m o l^{- 1}$