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 hard water and soft water heavy water study of heavy water hydrogen and its compounds

If a mole of hydrogen molecule is heated to a high temoerature then which of the following reactions take place?

  1. $H _2{(g)} + 436 kJ mol^{-1} \rightarrow H{(g)} + H{(g)}$
  2. $2H2{(g)} + 820 kJ mol^{-1} \rightarrow 2H _2{(g)}$
  3. $H _2{(g)} + H _2{(g)} + 436kJ mol^{-1} \rightarrow H^{+} _{(aq)} + H^{-} _{(aq)}$
  4. $H _2{(g)} + 200kJ mol^{-1} \rightarrow H _{(g)} + H _{(g)}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The amount of energy required to break $H-H$ bond of 1 mole of gaseous hydrogen is 436 $kJ \ mol^{-1}$. This is known as bond dissociation enthalpy.
$H _2{(g)} + 436 \  kJ mol^{-1} \rightarrow  H{(g)} + H{(g)}$

Multiple choice physics types of energy renewable and non-renewable resources renewable and non-renewable sources of energy substances, objects and energy

A body of mass 5 kg falls from a height of
30 metre. If its all mechanical energy is changed into heat, then heat produced
will be:-

  1. 350cal

  2. 150 cal

  3. 60cal

  4. 6cal
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
work done $=mgn$
$= 5 \times 9.8 \times 30 = 14705$
Heat produce $=\dfrac{1470}{4.2}= \boxed{350\ cal}$
$\boxed {Answer\ is\ A}$



























Multiple choice physics quantum physics photons concept of photon photons and photoelectric effect

Approximately, the temperature corresponding to $1 eV$ translational kinetic energy of molecule is

  1. $7.6 \times 10 ^ { 2 } \mathrm { K }$
  2. $7.7 \times 10 ^ { 3 } \mathrm { K }$
  3. $7.1 \times 10 ^ { - 2 } \mathrm { K }$
  4. $7.2 \times 10 ^ { 3 } \mathrm { K }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The translational kinetic energy of a molecule is given by E = (3/2)kT. Setting E = 1 eV = 1.6 * 10^-19 J, we solve for T = (2 * 1.6 * 10^-19) / (3 * 1.38 * 10^-23). This yields approximately 7727 K.

Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

Boron can undergo the following reactions with the given enthaly changes:
Assume no other reactions are occurring. If in a container (operating at constant pressure) which is isolated from the surrounding, mixture of are passed over excess of B(s), then calculate the molar ratio so that temperature of the container do not change :
${\text{2B}}\left( {\text{s}} \right){\text{ + }}\dfrac{{\text{3}}}{{\text{2}}}{{\text{O}} _{\text{2}}}\left( {\text{g}} \right) \to {{\text{B}} _{\text{2}}}{{\text{O}} _{\text{3}}}\left( {\text{s}} \right);\;\Delta H =  - 1260\;KJ$
${\text{2B}}\left( {\text{s}} \right){\text{ + 3}}{{\text{H}} _{\text{2}}}\left( {\text{g}} \right) \to {{\text{B}} _2}{H _6}\left( g \right);\;\Delta H = 30KJ$

  1. 15 : 3

  2. 42 : 1

  3. 1 : 42

  4. 1 : 84
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For the temperature of the isolated container to remain constant, the total enthalpy change for the reactions must be zero. Let x moles of B react to form B2O3 and y moles of B react to form B2H6. Using the given stoichiometry and enthalpy values, we solve for the ratio x:y to find the correct molar ratio.

Multiple choice physics heat energy transfers heat and heat transfer heat energy transfer transfer of heat

$\triangle { H }^{ \circ  }$ for reaction: 
${ F } _{ 2 }+2HCI\rightarrow 2HF+{ CI } _{ 2 }$
is equal to -352.8kJ. If $\triangle { H } _{ f }^{ \circ  }$ for HF is -268.3 kJ ${ mol }^{ -1 }$ then ${ \triangle H } _{ f }^{ \circ  }$ of HCI would be:- 

  1. -22 kJ ${ mol }^{ -1 }$
  2. 88.0 kJ ${ mol }^{ -1 }$
  3. 01.0 kJ${ mol }^{ -1 }$

  4. None
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

Bond dissociation enthalphies of ${ H } _{ 2 }\left( g \right) $ and ${ N } _{ 2\left( g \right)  }$ are enthalpy of formation of $N{ H } _{ 3 }\left( g \right) $ is $-46 kJ mol^{-1}$. What is enthalpy of atomization of $N{ H } _{ 3 }\left( g \right) $?

  1. $390.3 kJ mol^{-1}$
  2. $1170.9kJ mol^{-1}$
  3. $590 kJ mol^{-1}$
  4. $720 kJ mol^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice chemistry nitrogen and sulfur ammonia-properties and uses ammonia compounds of nitrogen - ammonia

The enthalpy of formation of ammonia is $-46.0KJ{mol}^{-1}$. The enthalpy change for the reaction 
$2{NH} _{3}(g)\rightarrow {N} _{2}(g)+3{H} _{2}(g)$

  1. $46.0KJ{mol}^{-1}$
  2. $92.0KJ{mol}^{-1}$
  3. $-23.0KJ{mol}^{-1}$
  4. $-92.0KJ{mol}^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The formation reaction of ammonia is N2(g) + 3H2(g) -> 2NH3(g) with an enthalpy change of -46.0 kJ per mole of ammonia formed. The reverse reaction given is 2NH3(g) -> N2(g) + 3H2(g), which reverses the sign and doubles the value since 2 moles are involved, yielding +92.0 kJ/mol.

Multiple choice chemistry nitrogen and sulfur ammonia-properties and uses ammonia compounds of nitrogen - ammonia

What is the amount of hear released when $3.4$ gm $NH _3$(g) and $6.4$gm $O _2$(g) react at constant temperature and pressure by the following equation.
$4NH _3$(g) + $5O _2$ (g) \rightarrow $4NO(g)$ + $6H _2O$(g) . $\Delta _1H^0 = 900 k.$

  1. $45$ kJ
  2. $250$ kJ
  3. $36$ kJ

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

The molecular weights of $\displaystyle NH _3$ and $\displaystyle O _2$ are 17 g/mol and 32 g/mol respectively.
3.4 g $\displaystyle NH _3 = \dfrac {3.4 \ g}{ 17 \ g/mol}=0.2 \ mol$
6.4 g $\displaystyle O _2= \dfrac {6.4 \ g}{32 \ g/mol}=0.2 \ mol$
0.2 moles of $\displaystyle O _2$ will react with $\displaystyle 0.2 \times \dfrac {4}{5}=0.16$ moles of $\displaystyle NH _3$. But 0.2 moles of
$\displaystyle NH _3$ are present. Hence, $\displaystyle O _2$ is the limiting reagent and $\displaystyle NH _3$ is excess reagent.

When 5 moles of $\displaystyle O _2$ react, the heat released is 900 kJ.
When 0.2 moles of $\displaystyle O _2$ react, the heat released will be
$\displaystyle 900 \ kJ \times \dfrac { 0.2}{5}=36 \ kJ$.

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 chemistry metals reaction of metals reactions of metals chemical properties of metals

The reaction of sodium metal with cold water is

  1. Endothermic

  2. Exothermic

  3. Both a and b

  4. Neither a nor b
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Metals like sodium and potassium react violently with cold water. These reactions are exothermic. The reaction is so violent and exothermic that the hydrogen which is evolved immediately catches fire.

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
$C _p=(4+f)R \ and \ C _v=(3+f)R$
$\therefore \dfrac{C _p}{C _v}=\dfrac{(4+f)R}{(3+f)R}=\dfrac{(4+f)}{(3+f)}$