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Calculate the lattice energy of KCl at 298 K. Given that:
Sublimation energy of K(s) = 89 kJ mol–1
Ionization energy of K(g) = 418 kJ mol–1
Dissociation energy of Cl2(g) = 244 kJ mol–1
Electro attachment energy of Cl(g) = –349 kJ mol–1
Formation of KCl(s) = – 437
Sublimation energy of K(s) = 89 kJ mol–1
Ionization energy of K(g) = 418 kJ mol–1
Dissociation energy of Cl2(g) = 244 kJ mol–1
Electro attachment energy of Cl(g) = –349 kJ mol–1
Formation of KCl(s) = – 437
- a)-575 kJ mol–1
- b)-202 kJ mol–1
- c)-717 kJ mol–1
- d)-800 kJ mol–1
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
Calculate the lattice energy of KCl at 298 K. Given that:Sublimation e...
-1
, ionization energy of K(g) = 418.8 kJ mol
-1
, electron affinity of Cl(g) = -349 kJ mol
-1
, and the distance between K+ and Cl- ions is 269 pm.
The lattice energy of KCl can be calculated using the Born-Haber cycle:
K(s) → K(g) ΔHsub = 89 kJ mol-1
K(g) → K+(g) + e- ΔHion = 418.8 kJ mol-1
Cl(g) + e- → Cl-(g) ΔHea = -349 kJ mol-1
K(s) + Cl(g) → KCl(s) ΔHlattice
The lattice energy can be calculated by summing up the enthalpy changes in the cycle:
ΔHlattice = -ΔHsub - ΔHion - ΔHea + ΔHf
where ΔHf is the enthalpy of formation of KCl, which can be assumed to be -436.8 kJ mol-1 at 298 K.
ΔHlattice = -89 kJ mol-1 - 418.8 kJ mol-1 - (-349 kJ mol-1) - 436.8 kJ mol-1
ΔHlattice = -703 kJ mol-1
To convert the enthalpy change to the lattice energy per mole of KCl, the Avogadro's number (Na) and the distance between K+ and Cl- ions (r) need to be considered:
U = ΔHlattice / Na * r
U = -703 kJ mol-1 / 6.022 x 1023 mol-1 * 269 pm * 1 m / 10^12 pm
U = -7156 kJ mol-1
Therefore, the lattice energy of KCl at 298 K is approximately -7156 kJ mol-1.
, ionization energy of K(g) = 418.8 kJ mol
-1
, electron affinity of Cl(g) = -349 kJ mol
-1
, and the distance between K+ and Cl- ions is 269 pm.
The lattice energy of KCl can be calculated using the Born-Haber cycle:
K(s) → K(g) ΔHsub = 89 kJ mol-1
K(g) → K+(g) + e- ΔHion = 418.8 kJ mol-1
Cl(g) + e- → Cl-(g) ΔHea = -349 kJ mol-1
K(s) + Cl(g) → KCl(s) ΔHlattice
The lattice energy can be calculated by summing up the enthalpy changes in the cycle:
ΔHlattice = -ΔHsub - ΔHion - ΔHea + ΔHf
where ΔHf is the enthalpy of formation of KCl, which can be assumed to be -436.8 kJ mol-1 at 298 K.
ΔHlattice = -89 kJ mol-1 - 418.8 kJ mol-1 - (-349 kJ mol-1) - 436.8 kJ mol-1
ΔHlattice = -703 kJ mol-1
To convert the enthalpy change to the lattice energy per mole of KCl, the Avogadro's number (Na) and the distance between K+ and Cl- ions (r) need to be considered:
U = ΔHlattice / Na * r
U = -703 kJ mol-1 / 6.022 x 1023 mol-1 * 269 pm * 1 m / 10^12 pm
U = -7156 kJ mol-1
Therefore, the lattice energy of KCl at 298 K is approximately -7156 kJ mol-1.
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Calculate the lattice energy of KCl at 298 K. Given that:Sublimation energy of K(s) = 89 kJ mol–1Ionization energy of K(g) = 418 kJ mol–1Dissociation energy of Cl2(g) = 244 kJ mol–1Electro attachment energy of Cl(g) = –349 kJ mol–1Formation of KCl(s) = – 437a)-575 kJ mol–1b)-202 kJ mol–1c)-717 kJ mol–1d)-800 kJ mol–1Correct answer is option 'C'. Can you explain this answer?
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Calculate the lattice energy of KCl at 298 K. Given that:Sublimation energy of K(s) = 89 kJ mol–1Ionization energy of K(g) = 418 kJ mol–1Dissociation energy of Cl2(g) = 244 kJ mol–1Electro attachment energy of Cl(g) = –349 kJ mol–1Formation of KCl(s) = – 437a)-575 kJ mol–1b)-202 kJ mol–1c)-717 kJ mol–1d)-800 kJ mol–1Correct answer is option 'C'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Calculate the lattice energy of KCl at 298 K. Given that:Sublimation energy of K(s) = 89 kJ mol–1Ionization energy of K(g) = 418 kJ mol–1Dissociation energy of Cl2(g) = 244 kJ mol–1Electro attachment energy of Cl(g) = –349 kJ mol–1Formation of KCl(s) = – 437a)-575 kJ mol–1b)-202 kJ mol–1c)-717 kJ mol–1d)-800 kJ mol–1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the lattice energy of KCl at 298 K. Given that:Sublimation energy of K(s) = 89 kJ mol–1Ionization energy of K(g) = 418 kJ mol–1Dissociation energy of Cl2(g) = 244 kJ mol–1Electro attachment energy of Cl(g) = –349 kJ mol–1Formation of KCl(s) = – 437a)-575 kJ mol–1b)-202 kJ mol–1c)-717 kJ mol–1d)-800 kJ mol–1Correct answer is option 'C'. Can you explain this answer?.
Calculate the lattice energy of KCl at 298 K. Given that:Sublimation energy of K(s) = 89 kJ mol–1Ionization energy of K(g) = 418 kJ mol–1Dissociation energy of Cl2(g) = 244 kJ mol–1Electro attachment energy of Cl(g) = –349 kJ mol–1Formation of KCl(s) = – 437a)-575 kJ mol–1b)-202 kJ mol–1c)-717 kJ mol–1d)-800 kJ mol–1Correct answer is option 'C'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Calculate the lattice energy of KCl at 298 K. Given that:Sublimation energy of K(s) = 89 kJ mol–1Ionization energy of K(g) = 418 kJ mol–1Dissociation energy of Cl2(g) = 244 kJ mol–1Electro attachment energy of Cl(g) = –349 kJ mol–1Formation of KCl(s) = – 437a)-575 kJ mol–1b)-202 kJ mol–1c)-717 kJ mol–1d)-800 kJ mol–1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the lattice energy of KCl at 298 K. Given that:Sublimation energy of K(s) = 89 kJ mol–1Ionization energy of K(g) = 418 kJ mol–1Dissociation energy of Cl2(g) = 244 kJ mol–1Electro attachment energy of Cl(g) = –349 kJ mol–1Formation of KCl(s) = – 437a)-575 kJ mol–1b)-202 kJ mol–1c)-717 kJ mol–1d)-800 kJ mol–1Correct answer is option 'C'. Can you explain this answer?.
Solutions for Calculate the lattice energy of KCl at 298 K. Given that:Sublimation energy of K(s) = 89 kJ mol–1Ionization energy of K(g) = 418 kJ mol–1Dissociation energy of Cl2(g) = 244 kJ mol–1Electro attachment energy of Cl(g) = –349 kJ mol–1Formation of KCl(s) = – 437a)-575 kJ mol–1b)-202 kJ mol–1c)-717 kJ mol–1d)-800 kJ mol–1Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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Here you can find the meaning of Calculate the lattice energy of KCl at 298 K. Given that:Sublimation energy of K(s) = 89 kJ mol–1Ionization energy of K(g) = 418 kJ mol–1Dissociation energy of Cl2(g) = 244 kJ mol–1Electro attachment energy of Cl(g) = –349 kJ mol–1Formation of KCl(s) = – 437a)-575 kJ mol–1b)-202 kJ mol–1c)-717 kJ mol–1d)-800 kJ mol–1Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Calculate the lattice energy of KCl at 298 K. Given that:Sublimation energy of K(s) = 89 kJ mol–1Ionization energy of K(g) = 418 kJ mol–1Dissociation energy of Cl2(g) = 244 kJ mol–1Electro attachment energy of Cl(g) = –349 kJ mol–1Formation of KCl(s) = – 437a)-575 kJ mol–1b)-202 kJ mol–1c)-717 kJ mol–1d)-800 kJ mol–1Correct answer is option 'C'. Can you explain this answer?, a detailed solution for Calculate the lattice energy of KCl at 298 K. Given that:Sublimation energy of K(s) = 89 kJ mol–1Ionization energy of K(g) = 418 kJ mol–1Dissociation energy of Cl2(g) = 244 kJ mol–1Electro attachment energy of Cl(g) = –349 kJ mol–1Formation of KCl(s) = – 437a)-575 kJ mol–1b)-202 kJ mol–1c)-717 kJ mol–1d)-800 kJ mol–1Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of Calculate the lattice energy of KCl at 298 K. Given that:Sublimation energy of K(s) = 89 kJ mol–1Ionization energy of K(g) = 418 kJ mol–1Dissociation energy of Cl2(g) = 244 kJ mol–1Electro attachment energy of Cl(g) = –349 kJ mol–1Formation of KCl(s) = – 437a)-575 kJ mol–1b)-202 kJ mol–1c)-717 kJ mol–1d)-800 kJ mol–1Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Calculate the lattice energy of KCl at 298 K. Given that:Sublimation energy of K(s) = 89 kJ mol–1Ionization energy of K(g) = 418 kJ mol–1Dissociation energy of Cl2(g) = 244 kJ mol–1Electro attachment energy of Cl(g) = –349 kJ mol–1Formation of KCl(s) = – 437a)-575 kJ mol–1b)-202 kJ mol–1c)-717 kJ mol–1d)-800 kJ mol–1Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Chemistry tests.
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