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Born-haber cycle - class-XI

Description: born-haber cycle
Number of Questions: 29
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Tags: chemistry thermodynamics chemical bonding and molecular structure energetics and thermochemistry lattice energy
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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


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?

  1. Sc

  2. Mn

  3. Cu

  4. Zn


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?

  1. $738 KJ $

  2. $ 10 KJ $

  3. $-371 kj$

  4. $-822 KJ$


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 $A + 2B \rightarrow 3C$?

  1. $-5$

  2. $+5$

  3. $+10$

  4. $+15$


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:


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

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

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

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

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

  1. $653.5\ kJ$

  2. $-686.2\ kJ$

  3. $-747.4\ kJ$

  4. None of these


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.

  1. 0

  2. 1

  3. 2

  4. 3


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.

  1. $39$

  2. $40$

  3. $41$

  4. None of these


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.

  1. 124 kJ/mol

  2. -124 kJ/mol

  3. 124 J/mol

  4. None of these


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

  1. $y$

  2. $y/2$

  3. $2y$

  4. $y/3$


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

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

  1. +640 kJ

  2. +1280 kJ

  3. -410 kJ

  4. +410 kJ


Correct Option: A

The lattice energy of NaCl(s) using the following data will be:

heat of sublimation of $Na(s)\,=\,S$

$(IE) _1$ of $Na\,(g)\,=\,I$

bond dissociation energy of $Cl _2\,(g)\,=\,D$

electron affinity of $Cl\,(g)\,=\,-E$

heat of formation of $NaCl(s)\,=\,-Q$

  1. Lattice energy $-U\, =\, S\, +\, I\, +\,\displaystyle \frac{D}{2}\, -\, E\, -\,Q$

  2. Lattice energy $-U\, =\, S\, -\, I\, +\,\displaystyle \frac{D}{2}\, -\, E\, -\,Q$

  3. Lattice energy $-U\, =\, S\, +\, I\, +\,\displaystyle \frac{D}{2}\, +\, E\, -\,Q$

  4. Lattice energy $-U\, =\, S\, -\, I\, -\,\displaystyle \frac{D}{2}\, +\, E\, +\,Q$


Correct Option: A
Explanation:

An estimate of the strength of the bonds in an ionic compound can be obtained by measuring the lattice energy of the compound, which is the energy given off when oppositely charged ions in the gas phase come together to form a solid.

The Lattice energy of NaCl from its elements Sodium and Chlorine in their stable forms is modeled in five steps in the diagram:

  1. Enthalpy change of atomization enthalpy of lithium
  2. Ionization enthalpy of lithium
  3. Atomization enthalpy of fluorine
  4. Electron affinity of fluorine
  5. Lattice enthalpy
$-U\, =\, S\, +\, I\, +\,\displaystyle \frac{D}{2}\, -\, E\, -\,Q$

Use the following data to calculate second electron ainity of oxygen, i.e., for the process
$O^{-}(g) + e^{-}(g)  \rightarrow O^{2-}(g)$
Is the $O^{2-}$ ion stable in the gas phase?.Why is it stable in solid MgO?
Heat of sublimation of $Mg(s) = + 147.7 kJ mol^{-1}$
Ionisation energy of Mg(g) to form
$Mg^{2+}(g) = + 2189.0 kJ mol^{-1}$
Bond dissociation energy for $O _2 = + 498.4 kJmol^{-1}$
First electron affinity of $O(g) = - 141.0 kJ mol^{-1}$
Heat formation of $MgO(s) = -601.7 kJ mol^{-1}$
Lattice energy of $MgO = -3791.0 kJ mol^{-1}$

  1. 601.7

  2. 744.4

  3. 1346.1

  4. 147.7


Correct Option: B
Explanation:

Option (B) is correct.

$\triangle H _1=-1346.1+q\ Mg(s)+\frac { 1 }{ 2 } O _{ 2 }(g)\rightarrow MgO(s)\ \triangle H _{ 2 }=-601.7\ By\quad Born-Haber\quad Cycle(based\quad on\quad Hess\quad law)\ \triangle H _{ 1 }=\triangle H _{ 2 }\ -1346.1+q=-601.7\ \qquad \qquad q=744.4kJ\quad mol^{-1}$

Select correct statement.

  1. Both lattice energy and hydration energies decrease with ionic size.

  2. Lattice energy can be calculated using Born-Haber cycle.

  3. If the anion is larger compared to the cation, the lattice energy will remain almost constant within a particular group.

  4. All the above are correct statements.


Correct Option: D
Explanation:

All the given statements are correct.
(A) Both lattice energy and hydration energies decrease with ionic size With large size, the charge is more dispersed. hence, the extent of hydration is smaller. Also the packing is inefficient.
(B) Lattice energy can be calculated using Born-Haber cycle The BornHaber cycle is an approach to analyze reaction energies. The cycle is concerned with the formation of an ionic compound from the reaction of a metal  (often a Group I or Group II element) with a halogen.
(C) If the anion is larger compared to the cation, the lattice energy will remain almost constant within a particular group

Caesium chloride is formed according to the following equation:

$Cs(s)+0.5{Cl} _{2}(g)\longrightarrow CsCl(s)$

The enthalpy of sublimation of $Cs$, enthalpy of dissociation of chlorine, ionization energy of $Cs$ and electron affinity of chlorine are $81.2, 243.0, 375.7$ and $-348.3kJ$ ${ol}^{-1}$. The energy change involved in the formation of $CsCl$ is $388.6\ kJ.{mol}^{-1}$. Calculate the lattice energy of $CsCl$.

  1. $-618.7\ kJ{mol}^{-1}$

  2. $+618.7\ kJ{mol}^{-1}$

  3. $1315.2\ kJ{mol}^{-1}$

  4. None of these


Correct Option: A

The ionization potential of ithium is 520 KJ/ mole .The energy required to convert 70 mg of lithum atoms in gaseous state into $Li^{+}$ ions is ______________.

  1. 52 KJ

  2. 5.2 KJ

  3. 520 KJ

  4. 52 J


Correct Option: B

The atoms of hydrogen combine to form a molecule of hydrogen gas, the energy of the $H _2$ molecule is:

  1. Greater than that of seperate atoms

  2. Equal to that of seperate atoms

  3. Lower than that of seperate atoms

  4. Some times lower and some times higher


Correct Option: C

Which of the following bonds has the highest bond energy?

  1. Si-Si

  2. Si=Si

  3. Si-O

  4. Si=O


Correct Option: D

Born Haber cycle is used to determine:

  1. electron affinity

  2. lattice energy

  3. crystal energy

  4. all the above


Correct Option: D

Choose the correct statement about the process (I) and (II).


$\displaystyle Na^{+} _{(g)}\xrightarrow{(I)}Na _{\left ( g \right )} \xrightarrow {(II)}Na _{\left ( s \right )}$

  1. In $(I)$ energy released. $(II)$ energy absorbed

  2. In both $(I)$ and $(II)$ energy is absorbed

  3. In both $(I)$ and $(II)$ energy is released

  4. In $(I)$ energy absorbed, $(II)$ energy released


Correct Option: C
Explanation:

The first process represents the electron affinity of $\displaystyle Na^+ $  ion which is negative.
The second process represents sublimation of $Na$. The sublimation energy is negative.
Thus, in both the processes, the energy is released.


Hence, the correct option is C.

For the maximum ionic character during bond formation ,_________ on cation and a __________ anion are required.

  1. low charge, high charge

  2. low charge, low charge

  3. high charge , no charge

  4. none of these


Correct Option: B
Explanation:
For an ionic interaction, both the cation and the anion should possess low charges. If the degree of polarization is quite small, ionic bonds are formed,
Otherwise, with higher degree of polarization, covalent bonds are formed
Ionic bonds result from a redox reaction in which the atoms of an element , which (usually metal) has low ionization energy, give away some of their electrons to achieve a stable electronic configuration

The standard formation reaction for aluminium oxide is $4Al(s) + 6O(g)\rightarrow 2Al _2O _3(s)$. This statement is false because:

  1. aluminium is not in its standard state

  2. the reaction should be written for one mole of Al

  3. The reaction is prevented by a thin film of $Al _2O _3$ formed on the surface of Al

  4. none of these


Correct Option: B
Explanation:
$2Al+3O _2\rightarrow Al _2O _3$

The enthalpy of the above reaction is enthalpy of formation as one mole of $Al _2O _3$ is forming from elemental aluminum and oxygen gas. 

In the balanced equation for combustion of 1 mole of butane, $C _4H _{10}(g)$, the coefficient of oxygen is:

  1. 5/2

  2. 9/2

  3. 5

  4. 13/2


Correct Option: D
Explanation:
Balanced chemical reaction of butane:
$C _4H _{10}+\frac{13}{2}O _2\rightarrow4CO _2+5H _2O$

Standard enthalpy of formation $(\Delta H _f)$ of which of the following is zero at $25^0C$ ?

  1. White phosphorous

  2. Red phosphorous

  3. Red lead $(Pb _3O _4)$

  4. $H^+(g)$


Correct Option: A
Explanation:

White phosphors is the elemental form of phosphorus for which enthalpy of formation is zero, while, $H^+$ and $Pb _3O _4$ are not elemental forms.

The enthalpy of hydrogenation for $1-pentene$ is $+126\ kJ/mol$. The enthalpy of hydrogenation for $1, 3-pentadiene$ is $+230\ kJ/mol$. Hence estimate the resonance magnitude of (delocalization) energy of $1, 3-pentadiene$.

  1. $22\, kJ/mol$

  2. $104\, kJ/mol$

  3. $252\, kJ/mol$

  4. cannot be calculated from this information


Correct Option: A
Explanation:

Enthalpy of hydrogenation per bond$=+126\ kJ/mol$


Enthalpy of hydrogenation for 2 bonds$=+252\ kJ/mol$


Resonance energy$=252-230=22\ kJ/mol$

The least stable in amongst the following is:

  1. $Li^-$

  2. $Be$

  3. $B$

  4. $C$


Correct Option: A
Explanation:

$Li$ is alkali metal hence, the negative charge on $Li$ is least stable as it will have the lowest electronegativity.

The heats of neutralization of $CH _3COOH.HCOOH, HCN$ and $HClO$ are 13.2, 13.4,2.9 and 3.6 kcal/eq respectively. Then, the degree of hydrolysis for the respective ions will be in the order :

  1. $CH _3COO^- < HCOO^- < CN^- < ClO^-$

  2. $HCOO^- < ClO^- < CN^- < CH _3COO^-$

  3. $CH _3COO^- < CN^- < ClO^- < HCOO^-$

  4. $HCOO^- < CH _3COO^- < ClO^- < CN^-$


Correct Option: D
Explanation:

Strong acid have high heat of neutralization and are very much stable so required high energy for dissociation. Hence, weak acids are hydrolysed at faster rate than strong acids.

The heat change for the reaction: $C(s) + S(s)\rightarrow CS _2(l)$, known as:

  1. heat of transition

  2. heat of fusion

  3. heat of vapourisation

  4. heat of formation


Correct Option: D
Explanation:
$C(s)+2S(s)\rightarrow CS _2(l)$
The heat change of the reaction is enthalpy of formation as 1 mole of $CS _2$ is being formed from elemental state of carbon as well as Sulphur.

Enthalpy of a solution of $ CsBr(s)$ is $10\ kJ/mol$.If the enthalpies of hydration of $Cs^+(g)$ and $Br(g)$ are 475 and 655 kJ.mol, what should be the lattice energy of $CsBr(s)$ in $kJ/mol$:

  1. $1120$

  2. $1130$

  3. $1140$

  4. $1150$


Correct Option: C
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