Question

$\\{ \huge{ \green .}}\bf \: \: \red{ H_{2(g)} \longrightarrow 2 H_{(g)} \:\:\:\:\:\triangle H = 436.08\: KJ}$

$\\{ \huge{ \green .}}\bf \: \: \red{ O_{2(g)} \longrightarrow 2 O_{(g)} \:\:\:\:\:\triangle H = 495.17\:KJ}$

$\\{ \huge{ \green .}}\bf \: \: \red{ H_{2(g)} + \dfrac{1}{2}O_{2(g)} \longrightarrow H_{2}O_{(g)} \:\:\:\:\:\triangle H = -241.84\: KJ}$

$\\{ \large{ \green \to}}\bf { \red{ \: \:\: Find \:\: bond \:\: energy \:\: of \:\: O-H \:\: bond.}}$​

Answer 1

ANSWER:

bond energy of single O-H bond is 462.7525 KJ

GIVEN:

$\\{ \huge{ \green .}}\bf \: \: \red{ H_{2(g)} \longrightarrow 2 H_{(g)} \:\:\:\:\:\triangle H = 436.08\: KJ}$..............1

$\\{ \huge{ \green .}}\bf \: \: \red{ O_{2(g)} \longrightarrow 2 O_{(g)} \:\:\:\:\:\triangle H = 495.17\:KJ}$.............2

$\\{ \huge{ \green .}}\bf \: \: \red{ H_{2(g)} + \dfrac{1}{2}O_{2(g)} \longrightarrow H_{2}O_{(g)} \:\:\:\:\:\triangle H = -241.84\: KJ}$..........3

TO FIND:

bond energy of O-H bond.

FORMULA:

∆H of reaction= bond energies of reactants - bond energy of product.

KEYS:

• ∆H of reaction is the energy released in equation 3

• bond energy of reactant includes energy of hydrogen (equation 1) and oxygen (equation 2).

• bond energy of two O-H bond can be represented as 2x .this represents the bond energy of product.

NOTE:

• there are two O-H bonds present in water.

• energy required to break the two O-H bonds is not same practically.

• but theoretically we can consider their bond energies approximately to be same.

IMPORTANT:

• bond energy are represented for single mole only.

• when 1/2 mole of oxygen is present ,the bond energy also becomes half

SOLUTION:

refer to the attachment

Answer 2

AnswEr :

Given,

• $\sf H_{2(g)} \longrightarrow 2 H_{(g)} \:\:\:\:\:\triangle H = 436.08\: KJ$

• $\sf O_{2(g)} \longrightarrow 2 O_{(g)} \:\:\:\:\:\triangle H = 495.17\:KJ$

• $\sf H_{2(g)} + \dfrac{1}{2}O_{2(g)} \longrightarrow H_{2}O_{(g)} \:\:\:\:\:\triangle H = -241.84\: KJ}$

We have to find the energy in O-H bonds.

We know that,

$\sf \Delta{}_ {rea}H =BE(reactants) - BE(products)$

Structure of Water :

• Water is a bent molecule with a bond angle of 104.5°
• A water molecule has two O-H bonds.

Let us assume the Bond Enthalpy of Water is 2x.

Therefore,

$\sf - 241.84 = \bigg(436.08 + \dfrac{495.17}{2} \bigg)- 2x \\ \\ \longrightarrow \sf \: - 241.84 = 683.665 - 2x \\ \\ \longrightarrow \sf \: 2x = 925.505 \\ \\ \longrightarrow \boxed{ \boxed{\sf \: x =462.75 \: kJ}}$

Thus,the bond energy in a O-H bond is 462.75 kJ