Chemistry

Chemical Equilibrium and Stoichiometry

135 Questions

Chemical equilibrium and stoichiometry questions cover equilibrium constants, percent yield, and the Haber process. These concepts test your ability to calculate reaction outcomes and purity. Practice these to excel in physical chemistry sections of competitive tests.

Equilibrium constant calculationsStoichiometric yieldHaber processPercent purityHomogenous equilibrium

Chemical Equilibrium and Stoichiometry Questions

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

if for the heterogeneous equilibrium $CaCO _{3}(s)\rightleftharpoons CaO(s)+CO _{2}(g);$ K=1 at 1 atm, the temperature is given by:

  1. $T=\frac{\Delta S^{0}}{\Delta H^{0}}$
  2. $T=\frac{\Delta H^{0}}{\Delta S^{0}}$
  3. $T=\frac{\Delta G^{0}}{ R^{0}}$
  4. $T=\frac{\Delta G^{0}}{\Delta H^{0}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Delta G = 2.303RT\space logK$


As K =1 , $\Delta G = 0$

We know the relation,

$\Delta G = \Delta H - T\Delta S$

$T = \dfrac{\Delta H}{\Delta S}$

Option B is correct

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

For the reaction : $\displaystyle 2NOCl(g)\longrightarrow 2NO(g)+{ Cl } _{ 2 }(g)$, The equilibrium constant at 400K, if $\displaystyle { \Delta H }^{ o }=77.18kJ{ mol }^{ -1 }$ and $\displaystyle { \Delta S }^{ o }=0.122kJ{ K }^{ -1 }{ mol }^{ -1 }$ is:

  1. $\displaystyle 1.97\times { 10 }^{ -3 }$
  2. $\displaystyle 1.97\times { 10 }^{ -2 }$
  3. $\displaystyle 1.97\times { 10 }^{ -4 }$
  4. $\displaystyle 1.97\times { 10 }^{ -1 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given the reaction: $2NOCl(g)\rightarrow 2NO(g)+Cl _2(g)$

$\Delta G^o=\Delta H^o-T\Delta S^o$

$\Delta G^o=77.18-400\times 0.122kJmol^{-1}$

$\Delta G^o=28.38\ kJmol^{-1}$

$K=e^{(\dfrac{-\Delta G^o}{RT})} $

$=1.97\times 10^{-4}$

Hence, option C is correct.
Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

For the equilibrium at $298$ K; $N _2O _4(g)\rightleftharpoons 2NO _2(g); G _{N _2O _4}^{\ominus}=100 kJ mol^{-1}$ and $G _{NO _2}^{\ominus}=50 kJ mol^{-1}$. If 5 mol of $N _2O _4$ and 2 moles of $NO _2$ are taken initially in one litre container than which statement are correct

  1. reaction proceeds in forward direction
  2. $K _c=1$
  3. $\Delta G=-0.55 kJ, \Delta G^{\ominus}=0$
  4. At equilibrium $[N _2O _4]=4.84 M$ and $[NO _2]=0.212 M$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

$\Delta G=\Delta G^{\ominus}+2.303 RT:log Q$

$\Delta G^{\ominus}=2\times G _{NO _2}^{\ominus}-G _{N _2O _4}^{\ominus}=2\times 50-100=0$
$\therefore \Delta G=0+2.303\times 8.314\times 10^{-3}\times 298: log \displaystyle\frac {22}{5}=0-0.55 kJ$

$\therefore \Delta G=-0.55 kJ$, i.e, reaction proceeds in forward direction
Also $\Delta G^{\ominus}=0=2.303 RT:log K \therefore K=1$
Now, $\underset {\underset {5-x}{5}}{N _2O _4}=\underset {\underset {2+2x}{2}}{2NO _2}$
$\therefore K _p=\frac {(P _{NO _2})}{(P _{N _2O _4})}=1=\frac {(2+2x)^2}{5-x}$ or  $x=0.106$


So, $[N _2O _4]=5-x=4.894M,\ [NO _2]=2+2x=2.12M$

Hence, options A, B and C are correct.

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

The density of an equilibrium mixture of $N _2O _4$ and $NO _2$ at 101.32 $KP _a$ is 3.62 g $dm^{3}$ at 288 K and 1.84 g $dm^{3}$ at 348 K. 


What is the heat of the reaction for the following reaction?

$N _2O _4\rightleftharpoons 2NO _2(g)$

  1. $\Delta _rH = 37.29 $ kJ mol$^{ -1 }$.
  2. $\Delta _rH = 75.68 $ kJ mol$^{ -1 }$.
  3. $\Delta _rH = 95.7$ kJ mol$^{ -1 }$.
  4. $\Delta _rH = 151.3 $ kJ mol$^{ -1 }$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

At 288 K, $M _{avg.}=\frac {3.62\times 0.0821\times 288}{1}$
$\frac {92}{M _{avg.}}=1+\alpha \Rightarrow K _{P1}=\frac {4\alpha^2}{1-\alpha^2}$
Similarly at $348 K, M'/avg.=\frac {1.84\times 0.0821\times 348}{1}$
$\frac {92}{M'avg}=1+\alpha'\Rightarrow K _{P _2}=\frac {4\alpha'^2}{1-\alpha'^2}$
$log \frac {K _{P _2}}{K _{P _1}}=\frac {\Delta H^o}{2.303 R}\left [\frac {1}{288}-\frac {1}{348}\right ]$
so,
$\Delta _rH = 75.68 kJ mol^{1}$

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

The cell in which the following reaction occurs:
$2Fe^{3+} _{(aq)}+2I^- _{(aq)}\rightarrow 2Fe^{2+} _{(aq)}+I _{2(s)}$ has $E^o _{cell}=0.236\ V$ at $298\ K$.
The equilibrium constant of the cell reaction is:

  1. $6.69\times 10^{-7}$
  2. $7.69\times 10^{-7}$
  3. $9.69\times 10^7$
  4. $6.69\times 10^7$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
We know
$\log K _c=\dfrac{nFE^0 _{cell}}{2.303RT}$
where,
$n=2$
$F=96487$
$E^0 _{cell}=0.236\ V$
$R=8.31$
$T=298$
Substituting the values, we get
$\log K _c=\dfrac{2\times96487\times0.236}{2.303\times8.31\times298}$
$\log K _c=7.9854$
$K _c=antilog (7.9854)$
$K _c=9.69\times10^7$
Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

$\Delta G^o (298 K)$ for the reaction $\dfrac12 N _2+\dfrac32H _2\overset {K _1}{\rightleftharpoons} NH _3$ is -16.5 kJ $mol^{-1}$. The equilibrium constant $(K _1)$ at $25^oC$ & the equilibrium constant $K _2$ and $K _3$ for the following reactions are
$N _2+3H _2\overset {K _2}{\rightleftharpoons} 2NH _3$
$NH _3\overset {K _3}{\rightleftharpoons } \dfrac12N _2+\dfrac32H _2$

  1. $K _1 = 779.4, K _2 = 6.074 \times 10^{5} ; K _3 = 1.283 \times 10^{-3}$
  2. $K _1 = 779.4, K _2 = 2.183 \times 10^{5} ; K _3 = 3.576 \times 10^{3}$
  3. $K _1 = 124.4, K _2 = 6.074 \times 10^{5} ; K _3 = 2.34\times 10^{3}$
  4. $None \:\:of \:\:these $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation


 $\displaystyle \Delta G^o = -RTlnK _1$
 $\displaystyle -16500 = - 8.314 \times 298 \times lnK _1$
$\displaystyle 6.6597 = ln K _1$
 $\displaystyle K _1 = 779.4$
$\displaystyle K _2 = K _1^2 = (779.4)^2 = 6.074 \times 10^5$
 $\displaystyle K _3 = \dfrac {1}{K _1}=\dfrac {1}{779.4}=1.283 \times 10^{-3} $

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

Calculate the equilibrium constant at 25 degrees celsius given the Standard Free Energy value of - 107.2 kJ

    • 43.2

  2. 43.2
  3. 6.18 x $ 10^8$

  4. 1.04
  5. 6.18 x $10^9$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 The temperature is 25 deg C or 298 K
$\displaystyle  \Delta G^0 = -RT lnK$
$\displaystyle  \Delta G^0 = - 107.2 kJ = -107200 J$
$\displaystyle  -107200 = - 8.314 \times 298 \times ln K$
$\displaystyle  ln K = 43.268$
$\displaystyle  K = 6.18 \times 10^{18}$

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

Consider the reaction of extraction of gold from its ore
$Au + 2CN^{-} (aq.) + \dfrac {1}{4}O _{2}(g) + \dfrac {1}{2}H _{2}O\rightarrow Au(CN) _{2}^{-} + OH^{-}$
Use the following data to calculate $\triangle G^{\circ}$ for the reaction
$K _{f} \left {Au(CN) _{2}^{-}\right ) = X$
$O _{2} + 2H _{2}O + 4e^{-}\rightarrow 4OH^{-}; E^{\circ} = +0.41\ volt$
$Au^{3+} + 3e^{-}\rightarrow Au; E^{\circ} = + 1.5\ volt$
$Au^{3+} + 2e^{-} \rightarrow Au^{+}; E^{\circ} = + 1.4\ volt$.

  1. $-RT\ ln\ X + 1.29\ F$
  2. $-RT\ ln\ X - 2.11\ F$
  3. $-RT\ ln \dfrac {1}{X} + 2.11\ f$
  4. $-RT\ ln\ X - 1.29\ F$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The standard Gibbs free energy change for the reaction is calculated using the standard cell potential. The reaction involves oxidation of Au and reduction of O2. The expression involves the formation constant X and the standard potentials provided.

Multiple choice chemistry mix and separate different types of mixtures homogenous and heterogenous mixtures all about mixtures

Mixture of $MgCO _{3}$ & $NaHCO _{3}$ on strong heating gives $CO _{2}$ & $H _{2}O$ in $3 : 1$ mole ratio. The weight $\%$ of $NaHCO _{3}$ present in the mixture is :

  1. $30\%$
  2. $80\%$
  3. $40\%$
  4. $50\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Heating MgCO3 gives CO2, and heating NaHCO3 gives CO2 and H2O. Let x be moles of MgCO3 and y be moles of NaHCO3. The total CO2 is x + 0.5y and H2O is 0.5y. Given (x + 0.5y) / (0.5y) = 3/1, we find x = y. Calculating the mass percentage based on molar masses (MgCO3=84, NaHCO3=84) leads to 50%.

Multiple choice rotational equilibrium option b: engineering physics motion of system of particles and rigid bodies equilibrium physics

If the potential energy of the molecule is given by $U = \dfrac {A}{r^{6}} - \dfrac {B}{r^{12}}$. Then at equilibrium position its potential energy is equal to

  1. $-A^{2}/ 4B$
  2. $A^{2}/ 4B$
  3. $2A/B$
  4. $A/2B$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

At equilibrium, the force F = -dU/dr = 0. Differentiating U = A/r^6 - B/r^12 gives -6A/r^7 + 12B/r^13 = 0, so r^6 = 2B/A. Substituting this back into U gives A/(2B/A) - B/(2B/A)^2 = A^2/2B - B * A^2/4B^2 = A^2/2B - A^2/4B = A^2/4B.

Multiple choice chemistry nitrogen and sulfur ammonia and fertilizers preparation of ammonia-laboratory method and haber's process ammonia

For the manufacture of ammonia by the reaction$N _{2}+3H _{2} \rightleftharpoons 2NH _{3}+21.9\ K cal$, the favorable conditions are:

  1. low temperature, low pressure & catalyst

  2. low temperature, high pressure & catalyst

  3. high temperature, low pressure & catalyst

  4. low temperature, high pressure

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

The equation representing habers process of ${ NH } _{ 3 }$

synthesis is-${ N } _{ 2 }(g)+{ 3H } _{ 2 }(g)\rightleftharpoons { 2NH } _{ 3 }(g)\ $
pressure: 150-200 atm
temperature : $450℃-500℃$
catalyst : Iron

Multiple choice chemistry nitrogen and sulfur ammonia and fertilizers preparation of ammonia-laboratory method and haber's process ammonia

$NH _3$ is formed in the following steps
I. $Ca+2C\rightarrow CaC _2$    50% yield
II. $CaC _2+N _2\rightarrow CaCN _2+C$   100% yield
III. $CaCN _2+3H _2O\rightarrow 2NH _3+CaCO _3$        50% yield
To obtain 2 moles $NH _3$, calcium required is:

  1. 1 mol

  2. 2 mol

  3. 3 mol

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

It can be easily observed from the reaction given that 1mole of Ca is required to get 2 moles of $NH _3$

Multiple choice chemistry nitrogen and sulfur ammonia and fertilizers preparation of ammonia-laboratory method and haber's process ammonia

Give the balanced reaction for the manufacture of the ammonia gas.

  1. $3{ N } _{ 2 }+3{ H } _{ 2 }\rightarrow 6{ NH } _{ 3 }$
  2. ${ N } _{ 2 }+3{ H } _{ 2 }\rightarrow 2{ NH } _{ 3 }$
  3. ${ N } _{ 2 }+3{ H } _{ 2 }\rightarrow { NH } _{ 3 }$
  4. ${ N } _{ 2 }+6{ H } _{ 2 }\rightarrow 4{ NH } _{ 3 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${ N } _{ 2 }+3{ H } _{ 2 }\rightarrow 2{ NH } _{ 3 }$
In the manufacture of ammonia process employed is Haber-Bosch which occurs in the presence of catalyst at high temperature and high pressure.