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Mazyrski [523]
2 years ago
7

Enthalpies of reaction calculated from bond energies and from enthalpies of formation are often, but not always, close to each o

ther.
(b) Ethylene glycol is produced by the catalytic oxidation of ethylene to ethylene oxide, which then reacts with water to form ethylene glycol:
The ΔH° for this hydrolysis step, based on enthalpies of formation, is - 97 kJ/mol. Calculate ΔH° for the hydrolysis using bond energies.
Chemistry
1 answer:
jolli1 [7]2 years ago
6 0

The enthalpy change in a reaction is given by-

ΔH°rxn = ∑nΔH°f,products - ∑nΔH°f,reactants

This can be expressed in terms of bond energy as-

ΔH°rxn = BEreactants - BEproducts

Therefore, the calculated bond energy according to the above equation will be-

ΔH°rxn = [ (C-C) + 2(C-O) + 4(C-H) + 2(O-H) ] - [ (C-C) + 2(C-O) + 4(C-H) + 2(O-H)  = 0 kJ/mol

<h3>What is enthalpy change?</h3>

Enthalpy change is a measure of the energy emitted or consumed in a reaction. This can be determined using the following equation which involves standard enthalpy of reactant and product formation:

ΔH°rxn = ∑nΔH°f,products - ∑nΔH°f,reactants

<h3>What is bond energy?</h3>

Bond energy is defined as the amount of energy needed to dissociate a mole of molecules into their individual atoms.

Learn more about the Enthalpy Change here:

brainly.com/question/14047927

#SPJ4

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18.2 mL of a 0.156 M solution of lead(II) nitrate are added to 26.2 mL of a 0.274 M solution of potassium iodide.
kow [346]

The mass of the Pbl2 : 1308.87

<h3>Further explanation</h3>

Given

18.2 mL of a 0.156 M Pb(NO3)2

26.2 mL of a 0.274 M KI

Reaction

Pb(NO3)2 (aq) + 2 KI (aq) - Pbl2 (s) + 2 KNO3

Required

the mass of the Pbl2

Solution

mol Pb(NO3)2 = 18.2 x 0.156 = 2.8392 mlmol

mol KI = 26.2 x 0.274 =7.1788 mol

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mol Pbl2 based on limiting reactant (Pb(NO3)2)

From equation, mol ratio of Pb(NO3)2 : Pbl2 = 1 : 1, so mol Pbl2=mol Pb(NO3)2=2.8392

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