Chemistry

Thermochemistry and Equilibrium

110 Questions

Thermochemistry and equilibrium problems focus on calculating bond energies, lattice enthalpy, and the Born Haber cycle. These concepts are vital for scoring well in chemistry sections. Regular practice ensures a clear understanding of energy changes in reactions.

Bond energy calculationsBorn Haber cycleLattice enthalpyEnthalpy of solutionThermochemical equations

Thermochemistry and Equilibrium Questions

Multiple choice chemistry chemical thermodynamics calorific value occurrence of carbon compounds in nature study of enthalpy

Calorific value of ethane, in kJ/g if for the reaction is: $2C _2H _6+7O _2\rightarrow 4CO _2+6H _2O;\Delta H=-745.6$ kcal.

  1. $-12.4$
  2. $-52$
  3. $-24.8$
  4. $-104$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Heat of combustion of ethane $=-\frac{745.6}{2}=-372.8\;Kcal\;mol^{-1}=-1559.79\;KJ\;mol^{-1}$       ($1\;cal.=4.2\;J$)

$calorific\; value=\dfrac{heat\;of\;combustion }{gram\;molecular\;weight\;of\;fuel}=-\dfrac{1559.79}{30}=-51.98\approx 52\;KJgm^{-1}$

Multiple choice chemistry chemical thermodynamics calorific value occurrence of carbon compounds in nature study of enthalpy

Calorific value of ethane, in kJ/g if for the reaction :


 $2C _2H _6 + 7O _2 \rightarrow 4CO _2 + 6H _2O; \Delta H = -745.6\ kcal$

  1. $-12.4$
  2. $-52$
  3. $-24.8$
  4. $-104$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$2C _2H _6+7O _2\rightarrow 4CO _2+6H _2O$; $\Delta H=-745.6KCal$


$\therefore$ Heat of combustion of ethane $=\cfrac {-745.6}{2}=-372.8 Kcal$

                                                                              $=-372.8 \times 4.18 kJ$

                                                                              $=-1558.3 kJ$

Calorific value $=\cfrac {\text{Heat of combustion}}{\text{Molecular weight of }C _2H _6}=- \cfrac {1558.3}{30}=-51.9=-52\ kJ/g$ .


Therefore, the correct option is B.

Multiple choice chemistry chemical thermodynamics calorific value occurrence of carbon compounds in nature study of enthalpy

The heat of combustion of carbon is $94$kcal. The calorific value of carbon is about:

  1. $7.8$ kcal
  2. $15.6$ kcal
  3. $47.0$ kcal
  4. $94$ kcal
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The calorific value of a substance is the heat which it releases during combustion of its $1\text{ } gm$ solid.

Therefore, $\text{Calorific value }(C.V)=\cfrac{\text{Heat of combustion}}{\text{Mol. weight}}\ C+O _2\longrightarrow CO _2;\triangle H=-94Kcal\C.V=\cfrac{-94}{12}Kcal\=-7.83Kcal$

Multiple choice evs how to save fuel fuels fuel and effects of burning fuel study of combustion

What is calorific value?

  1. Amount of heat energy produced on complete combustion of 1 Kg of fuel.

  2. Amount of heat energy produced on complete combustion of 100 Kg of fuel.

  3. Amount of heat energy produced on complete combustion of 1 g of fuel.

  4. Amount of heat energy lost on complete combustion of 1 Kg of fuel.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Calorific value is the quantity of heat produced by the complete combustion of a given mass of a fuel, usually expressed in joules per kilogram.

It can be defined as the amount of heat energy produced on complete combustion of 1 Kg of fuel. 


Hence, the correct option is $A$.


Multiple choice evs how to save fuel fuels fuel and effects of burning fuel study of combustion

Heats of combustion of $CH _4, C _2H _4, C _2H _6 $ are -890, -1411 and -1560 KJ/mole respectively. Which has the lowest fuel value in KJ/g?

  1. $CH _4$
  2. $C _2H _4$
  3. $C _2H _6$

  4. All same
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Fuel value $CH _4=\cfrac{890}{16}$ KJ/g=555.625 KJ/g

Fuel value $C _2H _4=\cfrac{1411}{28}$ KJ/g=50.4 KJ/g

Fuel value $C _2H _6=\cfrac{1560}{30}$ KJ/g=52KJ/mole

$\therefore $ Hence lowest fuel value for $C _2H _4$
Multiple choice chemistry how far? how fast? collision theory collision theory of chemical reactions rate of chemical reaction

In a reaction carried out at 400 k, $0.0001\%$ of the total number of collisions are effective. The energy of activation of the reaction is:

  1. zero

  2. 7.37 k cal/mol

  3. 9.212 k cal/mol

  4. 11.05 k cal/mol
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that, Arrhenius equation for calculation of energy of activation of reaction with rate constant $K$ and temeperature $T$ is 

$K=A$ $e^{-Ea/RT}$
where, $Ea$= Arrhenius activation energy
$A$= pre exponential factor (frequency factor)

Now, $e^{-Ea/K _BT}$= Fraction of collision having more than activation energy
where, $K _B=$ Boltzmann constant
Given, $T=400$ $K$       and effective collision= $0.0001$%

$\Rightarrow$ Effective Collision= $e^{-Ea/K _BT}$
$\Rightarrow$ $0.0001$%= $e^{-Ea/1.3\times 10^{-23}\times 400}$
$\Rightarrow$ $10^{-6}$= $e^{-Ea/1.3\times 10^{-23}\times 400}$

$\Rightarrow$ $2.303\times \log 10^{-6}$= $\cfrac {-Ea}{1.3\times 10^{-23}\times 400}$
$\Rightarrow$ $2.303\times (-6)$= $\cfrac {-Ea}{1.3\times 10^{-23}\times 400}$

$\Rightarrow$ $Ea$= $1.3\times 10^{-23}\times 400\times6\times 2.303=7.19\times 10^{-20}$ $J/mol$

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

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

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 : ${ \Delta H } _{ lattice }$ $KH = 710 kJ/mol$

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 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.

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

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 $-U\, =\, S\, -\, I\, -\,\displaystyle \frac{D}{2}\, +\, E\, +\,Q$
Reveal answer Fill a bubble to check yourself
A Correct answer
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
$-U\, =\, S\, +\, I\, +\,\displaystyle \frac{D}{2}\, -\, E\, -\,Q$

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Option (B) is correct.

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

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry lattice energy born-haber cycle born-haber cycles energy cycles

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. $H^+(g)$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.