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 bio-chemistry carbohydrate metabolism energy output living organisms and energy production respiration in cell
Amount of energy released during hydrolysis of a high energy bond of ATP is
  1. $73 Kcal \,{mol}^{-1}$
  2. $0.73 Kcal \,{mol}^{-1}$
  3. $3.4 Kcal\, {mol}^{-1}$
  4. $7.3 Kcal \,{mol}^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Adenosine triphosphate(ATP) is the energy currency of the cell. It is the most important energy carrier which carries energy in the two terminal phosphate bonds(called as high energy bonds or energy-rich bonds). Equal amounts of usable energy are released per mole of ATP or ADP hydrolysis:
$ATP+{H} _{2}O\rightleftharpoons ADP+Pi+7.3Kcal{\,\,mol}^{-1}$
$ADP+{H} _{2}O\rightleftharpoons AMP+Pi+7.3Kcal{\,\,mol}^{-1}$
So the correct answer is '$7.3 Kcal \,{mol}^{-1}$'.
Multiple choice

How much energy we can get from one ATP molecule?

  1. 2700 cal

  2. 7300 cal

  3. 7200 cal

  4. 1700 cal
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In cellular respiration, the hydrolysis of one ATP molecule into ADP and inorganic phosphate releases about 7200 calories (7.2 kcal) of usable energy according to standard school biology textbooks.

Multiple choice some important compounds of magnesium and calcium the s-block elements chemistry

Gypsum on heating to $300^{o}C$ gives:

  1. $CaSO _{4}2H _{2}O$
  2. $CaSO _{4}$
  3. $CaSO _{4}.\dfrac {1}{2}H _{2}O$
  4. $SO _{3}$ ans $CaO$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here answer is option C

Gypsum- chemical formula ;$Ca{ SO } _{ 4 }\cdot 2{ H } _{ 2 }O$
When gypsum is heated at 300K, we get plaster of paris(calcium sulphate hemihydrated salt)
$Ca{ SO } _{ 4 }\cdot 2{ H } _{ 2 }O\rightarrow Ca{ SO } _{ 4 }\frac { 1 }{ 2 } { H } _{ 2 }O+\frac { 3 }{ 2 } { H } _{ 2 }O$
Hence calcium sulphate Hemihydrated salt $Ca{ SO } _{ 4 }\frac { 1 }{ 2 } { H } _{ 2 }O$ is produced after heating gypsum.

Multiple choice chemistry matter in our surroundings diffusion in different states of matter properties of solids, liquid, and gas particle theory of matter

The translational kinetic energy of N molecules of $O _2$ is x J at $-123^o$C. Another sample of $O _2$ at $27^o$C has translational kinetic energy of $2x$ J. The latter sample contains:

  1. N molecules of $O _2$
  2. $2N$ molecules of $O _2$
  3. $N/2$ molecules of $O _2$
  4. $N/4$ molecules of $O _2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$T _1=-123^o$ K $=150$ K, $T _2=27^oC=300$ K


$KE _1=xJ$

$KE _2=2xJ$

Average translational kinetic energy:

$E=\dfrac{3}{2}nRT$

$=\dfrac{3}{2}\dfrac{N}{N _A}RT$

$\dfrac {x}{2x} = \dfrac{\dfrac{3NRT _1}{2N _A}}{\dfrac{3N _1RT _2}{2N _A}}$

$\dfrac{1}{2}=\dfrac{N}{N _1}\times \dfrac{150}{300}$

$N=N _1$

Hence, the correct answer is option $A$.

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

The equilibrium constants of a reaction is $73$. Calculate standard free energy change.

  1. $-106\ kJ\ mol^{-1}$
  2. $0.632\ kJ\ mol^{-1}$
  3. $60.32\ kJ\ mol^{-1}$
  4. $-10.632\ kJ\ mol^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

${ \triangle G }^{ 0 }=-RTlnK$             $T={ 27 }^{ 0 }C=300K$

${ \triangle G }^{ 0 }=-8.3\times 300\times ln73$
          $=-10.632KJ{ mol }^{ -1 }$

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

${ K } _{ C }$ for ${ 3 }/{ 2{ H } _{ 2 }+{ 1 }/{ 2{ N } _{ 2 }\rightleftharpoons  } }{ NH } _{ 3 }$ are 0.0266 and $0.0129\,{ atm }^{ -1 }\quad $ respectively, at 350$^o$C and 400$^o$C. Calculate the heat of formation of ${ NH } _{ 3 }$.

  1. $\therefore \triangle H =\,-50462\quad cal$
  2. $\therefore \triangle H=\,-8133\quad cal$
  3. $\therefore \triangle H =\,12140\quad cal$
  4. $\therefore \triangle H=\,-12140\quad cal$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As we know,
$2.303\,log[\displaystyle\frac { { K } _{ { P } _{ 2 } } }{ { K } _{ { P
} _{ 1 } } }] =\frac { \triangle H }{ R } \left[ \frac { { T } _{ 2 }-{ T
} _{ 1 } }{ { T } _{ 1 }{ T } _{ 2 } }  \right] $
$2.303\,log[\displaystyle\frac
{ 0.0129 }{ 0.0266 }] =\frac { \triangle H }{ 2 } \left[ \frac {
673-623 }{ 673\times 623 }  \right] $
$\therefore \triangle H=\,12140\quad cal\quad =\,-12140\quad cal$

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

Calculate the Standard Free Energy Change at 25 degrees celsius given the Equilibrium constant of 1.3 x 10^4.

  1. +23.4 kJ
    • 3.22 x 10^4 kJ

  3. -23,400 kJ

  4. -23.4 kJ

  5. +23,400 kJ
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

 The temperature is 25 deg C or 298 K.
$\displaystyle  \Delta G^o = -RT ln K = - 8.314 \times 298 \times ln 1.3 \times 10^4 = -23469 J = -23.4 kJ$
Hence, the standard free energy change is -23.4 kJ

Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

The total Kinetic energy of $1\ mole$ of ${N}^{} _{2}$ at $27^{o} _{}{C}$ will be approximately :-

  1. $1500\ J$
  2. $15633\ cal$
  3. $1500\ kcal$
  4. $1500\ erg$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For $n$ mole of any gas the total  kinetic energy is given as $E=\dfrac{3}{2}nRT$

Where $R$ is gas constant having value $8.31J/mole-K$ or $8.31\times 4.18 cal /mole-K=34.74\text{Cal per mole per Kelvin}$
$T$ is temperature in Kelvin which is $T=27+273=300K$
So putting all values we get $E=1.5\times 1 \times 34.74\times 300=15633Calorie$

Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

The total kinetic energy of $1$ mole of $N _2$ at $27$C will be approximately

  1. <span class="mrow"><span class="mn">3739.662 J

  2. 1500 calorie

  3. 1500 kilo calorie

  4. 1500 erg.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The kinetic enrgy of one mole is given by:

KE=$\dfrac{3}{2} K _BT$
The kinetic enrgy of 1 mole of $N _2$ atoms is:
KE=$\dfrac{3}{2}K _B T$ where $N$ is Avogadro's number,$K _B$ is Boltzmann's constant and $T$ is temperature
KE=$\dfrac{3}{2} \times (6.022 \times 10^{23})\times (1.38 \times 10^{-23}) \times 300$
$=3739.662 J$

Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

For polyatomic molecules having 'f' vibrational modes, the ratio of two specific heats, $\dfrac{C _p}{C _v}$ is ............

  1. $\dfrac{1+f}{2+f}$
  2. $\dfrac{2+f}{3+f}$
  3. $\dfrac{4+f}{3+f}$
  4. $\dfrac{5+f}{4+f}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

By the law of equipartition of energy, for one mole of polyatomic gas
$C _p=(4+f)R \ and \ C _v=(3+f)R$
$\therefore \dfrac{C _p}{C _v}=\dfrac{(4+f)R}{(3+f)R}=\dfrac{(4+f)}{(3+f)}$

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

The dissociation energy of $CH _4$ and $C _2H _6$ are respectively 360 and 620 kcal /mole. the bond energy $C-C$ is:

  1. 260 kcal /mole

  2. 180 kcal /mole

  3. 130 kcal / mole

  4. 80 kcal / mole
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Dissociation of energy of $CH _4$ base
=$\cfrac {\text{dissociation energy of }{CH _4}}{4}=\cfrac {360}{4}=90Kcal/mole$

Dissociation energy of $C _2H _6$

620=1(B.E of C-C bond+ 6 B.E of C-H bond)

620=B.E of C-C bond +6 $\times $90

B.E of C-C bond=620-540=80 kcal / mole
Multiple choice

What is the formula for chemical potential energy?

  1. $U = mgh$
  2. $U = rac{1}{2}mv^2$
  3. $U = rac{1}{2}kx^2$
  4. $U = - rac{k}{r}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Chemical potential energy is the energy an object possesses due to its chemical composition. The formula for chemical potential energy is $U = -rac{k}{r}$, where $k$ is a constant and $r$ is the distance between the particles.

Multiple choice

What is the potential energy of a chemical reaction?

  1. $U = mgh$
  2. $U = rac{1}{2}mv^2$
  3. $U = rac{1}{2}kx^2$
  4. $U = - rac{k}{r}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The potential energy of a chemical reaction is $U = -rac{k}{r}$, where $k$ is a constant and $r$ is the distance between the particles.