Chemistry

Thermochemistry and Equilibrium

121 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 energy fuel cell fuel cells electrochemistry, rechargeable batteries, and fuel cells

If a fuel cell methanol is used as fuel and oxygen gas is used as an oxidiser. The reaction is $CH _3OH _{(l)} +\frac{3}{2}O _{2(g)} \rightarrow CO _{2(g)}+2H _2O _{(g)}$ at 298 K standard Gibb's energies of formation for $CH _3OH(l)H _2O(l)$ and $CO _2(g)$ are $-166.2$, $-237.2$ and $394.4$ $KJ \ mol^{-1}$ respectively. If standard enthalpy of combustion of methanol is -726 KJ $mol^{-1}$, efficieny of the fuel cell will be 

  1. 80%

  2. 87%

  3. 90%

  4. 97%
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry chemical equilibrium relationship between equilibrium constant, reaction quotient and gibbs energy law of mass action chemical equilibrium and acids-bases

The equilibrium constant of the reaction $2C _3H _6 (g) \rightleftharpoons C _2H _4 (g) + C _4H _8 (g)$ is found to fit the expression:
                         $lnK=-1.04-\dfrac {1088}{T}$


Calculate the standard reaction enthalpy and entropy at 400 K :

  1. $\Delta H^o = 4.5 kJ/mol ; \Delta S^o = 4.32 J / mol^{1} K^{1}$
  2. $\Delta H^o = 9.04 kJ/mol ; \Delta S^o =- 8.64 J / mol^{1} K^{1}$
  3. $\Delta H^o = 18.08 kJ/mol ; \Delta S^o = 17.28 J / mol^{1} K^{1}$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\displaystyle  \Delta G^0 = -RTlnK =\Delta H^0 - T\Delta S^0$

$\displaystyle  lnK = \frac {\Delta S^0}{R}-\frac {\Delta H^0 }{RT} $......(1) 

But $\displaystyle lnK=-1.04-\frac {1088}{T} $......(2)

From (1) and (2),

$\displaystyle  \frac {\Delta S^0}{R} = -1.04 $ and $\displaystyle \frac {\Delta H^0 }{RT} =  \frac {1088}{T}$

$\displaystyle  \Delta S^0 = -1.04 \times 8.314 = - 8.64 J/mol/K$

$\displaystyle \Delta H^0  = 1088 \times 8.314 =9040 J/mol =9.04 kJ/mol $
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

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
${ \Delta H } _{ r }$= $- 124 kJ/mole$