Tag: born-haber cycle

Questions Related to born-haber cycle

Which of the following can be calculated from Born-Haber cycle for $Al _2O _3$?

chemistry lattice energy born-haber cycle born-haber cycles energy cycles

  1. Lattice energy of $Al _2O _3$

  2. Electron affinity of O-atom

  3. Ionisation energy of Al

  4. All of these

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Correct Option: D
Explanation:

 The Born-Haber Cycle can be applied to determine the lattice energy of an ionic solid; ionization energy, electron affinity, dissociation energy, sublimation energy, heat of formation, and Hess's Law.

In the series $Sc(Z=21)$ to $Zn(Z=30)$, the enthalpy of atomisation of which element is least?

chemistry lattice energy born-haber cycle born-haber cycles energy cycles

  1. Sc

  2. Mn

  3. Cu

  4. Zn

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Correct Option: D
Explanation:

$Sc\  \& \ Zn$ belongs to the third group of the periodic table. The extent of metallic bonding an element undergoes decides the enthalpy of atomization. The more extensive the metallic bonding of an element, the more will be its enthalpy of atomization. In all transition metals, there are some unpaired electrons that account for their stronger metallic bonding. 

Due to the absence of these unpaired electrons, the inter-atomic electronic bonding is the weakest in $Zn$ and as a result, it has the least enthalpy of atomization.

The lattice energy of CsI(s) is −604 KJ/mol, and the enthalpy of solution is 33 KJ/mol. How would you calculate the enthalpy of hydration (KJ) of 0.65 moles of CSI? Enter a numeric answer only, do not include units in your answer?

chemistry lattice energy born-haber cycle born-haber cycles energy cycles

  1. $738 KJ $

  2. $ 10 KJ $

  3. $-371 kj$

  4. $-822 KJ$

Show answer
Correct Option: C

Consider the following reaction,
$2A + B \rightarrow C + 2D$, $\Delta H _{1} = 10$
$A + 2C \rightarrow 2D + B$, $\Delta H _{2} = -5$ What is $\Delta H$ of reaction $A + 2B \rightarrow 3C$?

chemistry lattice energy born-haber cycle born-haber cycles energy cycles

  1. $-5$

  2. $+5$

  3. $+10$

  4. $+15$

Show answer
Correct Option: D
Explanation:
Solution:- (D) $+15$
$A + 2B \longrightarrow 3C \quad \Delta{H} = ?$
Given:-
$2A + B \longrightarrow C + 2 D \quad \Delta{{H} _{1}} = 10 ..... \left( 1 \right)$
$A + 2C \longrightarrow 2D + B \quad \Delta{{H} _{2}} = -5$
$B + 2D \longrightarrow A + 2C \quad \Delta{{H} _{3}} = - \Delta{{H} _{2}} = 5 ..... \left( 2 \right)$
Adding ${eq}^{n} \left( 1 \right) \& \left( 2 \right)$, we have
$2A + B + B + 2D \longrightarrow C + 2D + A + 2C \quad \Delta{H} = 10 + 5$
$A + 2B \longrightarrow 3C \quad \Delta{H} = 15$
Hence the $\Delta{H}$ for the given reaction is $+15$.

Determine ${ \Delta  }{ U }^{ o }$ at $300K$ for the following reaction using the listed enthalpies of reaction:


$4CO(g)+8{ H } _{ 2 }(g)\longrightarrow 3{ CH } _{ 4 }(g)+{ CO } _{ 2 }(g)+2{ H } _{ 2 }O(l)$

$C _{(graphite)}+1/2{ O } _{ 2 }(g)\longrightarrow CO(g);\quad \Delta { { H } _{ 1 } }^{ o }=-110.5kJ$

$CO(g)+1/2{ O } _{ 2 }(g)\longrightarrow { CO } _{ 2 }(g);\quad \Delta { { H } _{ 2 } }^{ o }=-282.9kJ$

${ H } _{ 2 }(g)+1/2{ O } _{ 2 }(g)\longrightarrow { H } _{ 2 }O(l);\quad \Delta { { H } _{ 3 } }^{ o }=-285.8kJ$

$C _{(graphite)}+2{ H } _{ 2 }(g)\longrightarrow { CH } _{ 4 }(g);\quad \Delta { { H } _{ 4 } }^{ o }=-74.8kJ$

chemistry lattice energy born-haber cycle born-haber cycles energy cycles

  1. $653.5\ kJ$

  2. $-686.2\ kJ$

  3. $-747.4\ kJ$

  4. None of these

Show answer
Correct Option: D

The Born Haber cycle below represents the energy changes occurring at 298K when KH is formed from its elements
v : ${ \Delta H } _{ atomisation }$ K = 90 kJ/mol
w : ${ \Delta H } _{ ionisation }$ K = 418 kJ/mol
x : ${ \Delta H } _{ dissociation }$ H = 436 kJ/mol
y : ${ \Delta H } _{ electron affinity }$ H = 78 kJ/mol
z : ${ \Delta H } _{ lattice }$ KH = 710 kJ/mol

In terms of the letters v to z the expression for
${ \Delta H } _{ i }$ of K is ${ \Delta H } _{ i }$ = $w/2$.
If true enter 1, else enter 0.

chemistry lattice energy born-haber cycle born-haber cycles energy cycles

  1. 0

  2. 1

  3. 2

  4. 3

Show answer
Correct Option: A
Explanation:

In terms of the letters v to z the expression for
${ \Delta H } _{ i }$ of K is ${ \Delta H } _{ i }$ = $w$.

The Born Haber cycle below represents the energy changes occurring at 298K when $KH$ is formed from its elements
v : ${ \Delta H } _{ atomisation }$ $K = 90 kJ/mol$
w : ${ \Delta H } _{ ionisation }$ $K = 418 kJ/mol$
x : ${ \Delta H } _{ dissociation }$ $H = 436 kJ/mol$
y : ${ \Delta H } _{ electron affinity }$ $H = 78 kJ/mol$
z : ${ \Delta H } _{ lattice }$ $KH = 710 kJ/mol$
On complete reaction with water, $0.1 g$ of $KH$ gave a solution requiring 25 ${ cm }^{ 3 }$ of 0.1M $HCl$ for neutralisation.Calculate the relative atomic mass of potassium from this information.

chemistry lattice energy born-haber cycle born-haber cycles energy cycles

  1. $39$

  2. $40$

  3. $41$

  4. None of these

Show answer
Correct Option: A
Explanation:

Meq. of KH = Meq. of $HCl$
$\frac { { 0.1 } }{ { E } _{ KH } } \times 1000\quad =\quad 25\quad \times \quad 0.1$
Valency factor (05955) of $K$ is 1 hence
${ E } _{ K }$=${ M } _{ K }$                    ${ M } _{ K }$=39
${ E } _{ KH }$=40                                      ${ E } _{ KH }$=${ E } _{ K }$=
40=${ E } _{ K }$+1                                     ${ E } _{ KH }$
${ E } _{ K }$ $\Rightarrow $ 39

The Born Haber cycle below represents the energy changes occurring at 298K when $KH$ is formed from its elements
v : ${ \Delta H } _{ atomisation }$ $K = 90 kJ/mol$
w : ${ \Delta H } _{ ionisation }$ $K = 418 kJ/mol$
x : ${ \Delta H } _{ dissociation }$ $H = 436 kJ/mol$
y : ${ \Delta H } _{ electron affinity }$ $H = 78 kJ/mol$
z : ${ \Delta H } _{ lattice }$ $KH = 710 kJ/mol$

Calculate the value of $\Delta $$H$ showing all your working.

chemistry lattice energy born-haber cycle born-haber cycles energy cycles

  1. 124 kJ/mol

  2. -124 kJ/mol

  3. 124 J/mol

  4. None of these

Show answer
Correct Option: B
Explanation:

Born-Haber Cycle

It is a series of steps (chemical processes) used to calculate the lattice energy of ionic solids, which is difficult to determine experimentally. You can think of BH cycle as a special case of  Hess's law which states that the overall energy change in a chemical process can be calculated by breaking down the process into several steps and adding the energy change from each step.

${ \Delta H } _{ r }$=2 $\times $ 90 + 2 $\times $ 418 + 436 - 2 $\times $ 78 - 2 $\times $ 710
${ \Delta H } _{ r }$= $- 124 kJ/mole$

The Born Haber cycle below represents the energy changes occurring at 298K when KH is formed from its elements
v : ${ \Delta H } { atomisation }$ K = 90 kJ/mol
w : ${ \Delta H } _{ ionisation }$ K = 418 kJ/mol
x : ${ \Delta H } _{ dissociation }$ H = 436 kJ/mol
y : ${ \Delta H } _{ electron affinity }$ H = 78 kJ/mol
z : ${ \Delta H } _{ lattice }$ KH = 710 kJ/mol

In terms of the letters v to z the expression for
${ \Delta H } _{ electron affinity }$ of H is ${ \Delta H } _{ electron affinity }$  is _.
I

chemistry lattice energy born-haber cycle born-haber cycles energy cycles

  1. $y$

  2. $y/2$

  3. $2y$

  4. $y/3$

Show answer
Correct Option: A
Explanation:
The electron affinity $\Delta { H } _{ electron\: affinity }$  of an atom or molecule is defined as the amount of energy  released when an electron is added to a neutral atom or molecule in the gaseous state to form a negative ion. It is exothermic as heat energy is released from the system.

$\Delta { H } _{ electron\: affinity }$ should be y not y/2.

The energy change for the alternating reaction that yields chlorine sodium $(Cl^{+}Na^{-})$ will be:

$2Na(s)\, +\, Cl _2(g)\,\rightarrow\, 2Cl^{+}Na^{-}(s)$

Given that:

Lattice energy of $NaCl\,=\,-787\, kJ\,mol^{-1}$

Electron affinity of $Na\,=\,-52.9\, kJ\, mol^{-1}$

Ionisation energy of $Cl\, =\, +\,1251\, kJ\, mol^{-1}$

BE of $Cl _2\,=\,244\, kJ\, mol^{-1}$

Heat of sublimation of $Na(s)\, =\,107.3\, kJ\, mol^{-1}$

$\Delta H _f(NaCl)\, =\,-411\, kJ\, mol^{-1}$.

chemistry lattice energy born-haber cycle born-haber cycles energy cycles

  1. +640 kJ

  2. +1280 kJ

  3. -410 kJ

  4. +410 kJ

Show answer
Correct Option: A