(A) For the reaction: $2 H_2 + O_2 \rightarrow 2 H_2 O$ (which contains $4 \times O-H$ bonds).$\Delta_{r} H = [2 \times BE(H-H) + BE(O=O)] - [4 \times

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(A) For the reaction: $2 H_2 + O_2 \rightarrow 2 H_2 O$ (which contains $4 \times O-H$ bonds).$\Delta_{r} H = [2 \times BE(H-H) + BE(O=O)] - [4 \times BE(O-H)]$$-571 = [2 \times 435 + 498] - 4 \times BE(O-H)$$-571 = [870 + 498] - 4 \times BE(O-H)$$-571 = 1368 - 4 \times BE(O-H)$$4 \times BE(O-H) = 1368 + 571$$4 \times BE(O-H) = 1939$$BE(O-H) = \frac{1939}{4} \approx 484.75 \ kJ/mol$Thus,the average bond energy is approximately $484 \ kJ/mol$.

Class 11 Chemistry Thermodynamics Heat of reaction, Bond energy and Hess law

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For the reaction $2 H_2 + O_2 \rightarrow 2 H_2 O$,$\Delta H = -571 \ kJ$. Bond energy of $H-H = 435 \ kJ$ and $O=O = 498 \ kJ$. Then the average bond energy of $O-H$ bond will be:

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A
$484 \ kJ/mol$
B
$-484 \ kJ/mol$
C
$271 \ kJ/mol$
D
$-271 \ kJ/mol$

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Thermodynamics

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Heat of reaction, Bond energy and Hess law

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For the reaction $2 H_2 + O_2 \rightarrow 2 H_2 O$,$\Delta H = -571 \ kJ$. Bond energy of $H-H = 435 \ kJ$ and $O=O = 498 \ kJ$. Then the average bond energy of $O-H$ bond will be:

Similar Questions

Given below are two statements:Assertion $(A)$: The enthalpy of formation of graphite is taken as zero.Reason $(R)$: Graphite is the thermodynamically most stable allotrope of carbon.The correct answer is:

In the following reaction,how much enthalpy change in $kJ$ is associated with the formation of $5.67 \ mol$ of $HCl$ gas? $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)} \; ; \Delta H = -184.6 \ kJ$

Which of the following reactions can be used to define the standard enthalpy of formation of $CO_{2_{(g)}}$?

The enthalpies of combustion of carbon and carbon monoxide are $-393.5 \, kJ$ and $-283 \, kJ$ respectively. Calculate the enthalpy of formation of carbon monoxide in $kJ$. (in $.5$)

The bond energy of $Cl-Cl$ in $Cl_2$ is $242 \, kJ \, mol^{-1}$. What is the wavelength (in $nm$) of light required to break a single $Cl-Cl$ bond?

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Given below are two statements:
Assertion $(A)$: The enthalpy of formation of graphite is taken as zero.
Reason $(R)$: Graphite is the thermodynamically most stable allotrope of carbon.
The correct answer is: