Question

refer to the attachment for the Question..


class 11 chemistry

kindly don't spam__/¡\__​

Answer 1

Answer:

use the answer attached hope it helps...

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the concept of Reaction Enthalpy has been used. The standard enthalpy change in the reaction is given as ∆H⁰ᵣ . This can be taken in any term. That is bond enthalpy in the reaction or molar enthalpies. Here we will be using the bond energies.

Let's do it !!

______________________________________

★ Formula Used :-

$\\\;\boxed{\displaystyle{\sf{\Delta H_{r} ^{o}\;\;=\;\bf{\sum_{i}\;a_{i}\:\Delta_{f}\:H_{(products)} ^{o}\;-\;\sum_{i}\;b_{i}\:\Delta_{f}\:H_{(reactants)} ^{o}}}}}$

______________________________________

★ Solution :-

Given,

» Reaction Enthalpy = ∆H⁰ᵣ = -2.05 × 10³ kJ/mol

» C - C bond energy = 347 kJ/mol

» C - H bond energy = 414 kJ/mol

» C = O bond energy = 741 kJ/mol

» O - H bond energy = 464 kJ/mol

We need to find the O = O bond energy in O₂ molecule. Then,

$\\\;\;\displaystyle{\sf{:\rightarrow\;\;\Delta H_{r} ^{o}\;\;=\;\bf{\sum_{i}\;a_{i}\:\Delta_{f}\:H_{(products)} ^{o}\;-\;\sum_{i}\;b_{i}\:\Delta_{f}\:H_{(reactants)} ^{o}}}}$

• Here, aᵢ and bᵢ are the stoichiometric coefficients of products and reactants respectively.
• Here ∆ H⁰ is the standard change in enthalpies of reactants and products.
• C = O bonds in products = 6
• O - H bonds in products = 8
• C - H bonds in reactants = 8
• C - C bonds in reactants = 2
• O = O bonds in reactants = 5

Applying values here, we get

$\\\;\;\displaystyle{\sf{:\Longrightarrow\;\;\Delta H_{r} ^{o}\;\;=\;\bf{[6(H_{C\:=\:O})\;+\;8(H_{O\;-\;H})]\;-\;[8(H_{C\;-\;H})\;+\;2(H_{C\;-\;C})\;+\;5(H_{O\;=\;O})]}}}$

Here H with subscripts shows the bond energies.

$\\\;\;\displaystyle{\sf{:\Longrightarrow\;\;-2.05\;\times\;10^{3}\;\;=\;\bf{[6(741)\;+\;8(464)]\;-\;[8(414)\;+\;2(347)\;+\;5(H_{O\;=\;O})]}}}$

$\\\;\;\displaystyle{\sf{:\Longrightarrow\;\;-2.05\;\times\;10^{3}\;\;=\;\bf{(4446\;+\;3712)\;-\;[3312\;+\;694\;+\;5(H_{O\;=\;O})]}}}$

$\\\;\;\displaystyle{\sf{:\Longrightarrow\;\;-2.05\;\times\;10^{3}\;\;=\;\bf{(4446\;+\;3712)\;-\;[3312\;+\;694\;+\;5(H_{O\;=\;O})]}}}$

$\\\;\;\displaystyle{\sf{:\Longrightarrow\;\;-2.05\;\times\;10^{3}\;\;=\;\bf{(8158)\;-\;[4006\;+\;5(H_{O\;=\;O})]}}}$

$\\\;\;\displaystyle{\sf{:\Longrightarrow\;\;-2050\;\;=\;\bf{8158\;-\;4006\;-\;5(H_{O\;=\;O})}}}$

$\\\;\;\displaystyle{\sf{:\Longrightarrow\;\;5(H_{O\;=\;O})\;\;=\;\bf{4152\;+\;2050}}}$

$\\\;\;\displaystyle{\sf{:\Longrightarrow\;\:\;5(H_{O\;=\;O})\;\;=\;\bf{4152\;+\;2050}}}$

$\\\;\;\displaystyle{\sf{:\Longrightarrow\;\:\;5(H_{O\;=\;O})\;\;=\;\bf{6202}}}$

$\\\;\;\displaystyle{\sf{:\Longrightarrow\;\;\:H_{O\;=\;O}\;\;=\;\bf{\dfrac{6202}{5}}}}$

$\\\;\;\displaystyle{\sf{:\Longrightarrow\;\;\:H_{O\;=\;O}\;\;=\;\bf{\green{1240.4\;\;kJ\:mol^{-1}}}}}$

$\\\;\underline{\boxed{\tt{Energy\;\;of\;\;O\;=\;O\;bond\;=\;\bf{\blue{1240.4\;\;kJ\:mol^{-1}}}}}}$

______________________________________

★ More Formulas to know :-

$\\\;\displaystyle{\sf{\leadsto\;\;W\;=\;-\:\int_{V_{i}} ^{V_{f}}\;p_{ex}\:dV}}$

$\\\;\displaystyle{\sf{\leadsto\;\;\Delta\;U\;=\;q\;+\;W}}$

$\\\;\displaystyle{\sf{\leadsto\;\;\Delta\;H\;=\;\Delta\:U\;+\;\Delta\:n_{g}\:RT}}$

$\\\;\displaystyle{\sf{\leadsto\;\;q\;=\;C\;Delta\:T}}$

$\\\;\displaystyle{\sf{\leadsto\;\;C_{p}\;-\;C_{v}\;=\;R}}$