Tag: born-haber cycle

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

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

  1. Lattice energy of $Al _2O _3$

  2. Electron affinity of O-atom

  3. Ionisation energy of Al

  4. All of these
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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

  1. Sc

  2. Mn

  3. Cu

  4. Zn
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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?

  1. $738 KJ $
  2. $ 10 KJ $
  3. $-371 kj$
  4. $-822 KJ$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice chemistry lattice energy born-haber cycle born-haber cycles energy cycles

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 $A + 2B \rightarrow 3C$?

  1. $-5$
  2. $+5$
  3. $+10$
  4. $+15$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $+15$.
Multiple choice chemistry lattice energy born-haber cycle born-haber cycles energy cycles

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


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

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

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

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

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

  1. $653.5\ kJ$
  2. $-686.2\ kJ$
  3. $-747.4\ kJ$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice chemistry lattice energy born-haber cycle born-haber cycles energy cycles

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.

  1. 0

  2. 1

  3. 2

  4. 3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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.

  1. $39$
  2. $40$
  3. $41$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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.

  1. 124 kJ/mol

  2. -124 kJ/mol

  3. 124 J/mol

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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
${ \Delta H } _{ r }$= $- 124 kJ/mole$

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

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

  1. $y$
  2. $y/2$
  3. $2y$
  4. $y/3$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
Multiple choice chemistry lattice energy born-haber cycle born-haber cycles energy cycles

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

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

  1. +640 kJ

  2. +1280 kJ

  3. -410 kJ

  4. +410 kJ
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The Born-Haber cycle for the formation of 2Cl+Na- involves sublimation of Na, dissociation of Cl2, ionization of Cl, electron affinity of Na, and lattice energy. Summing these steps to match the formation enthalpy of 2 units of the product gives the energy change.