Chemistry

Chemical Equilibrium and Stoichiometry

103 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

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 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

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.

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

Consider the equilibrium $NH _{4}Cl (s)\rightleftharpoons NH _{3}(g) + HCl(g)$. An inert gas is added to the system at constant volume and temperature. The correct statement(s) is/ are:

  1. the partial pressures of $NH _{3}$ and $HCl$ in the system will increase
  2. the partial pressures of $NH _{3}$ and $HCl$ in the system will remain the same
  3. the partial pressures of $NH _{3}$ and $HCl$ in the system will decrease

  4. the entropy of the system will increase
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation

$NH _4$ $+$ $Cl(s)$ $\rightleftharpoons$ $NH _3 (g)$ $+$ $HCl$.

Noble gas are  inert hence adding inert gas doesn't effect the partial pressure.
This is because noble gas gets added to both reactant and product.
$X (g)$ $+$ $NH _4Cl (S)$ $ \rightleftharpoons $ $NH _3(g)$ $+$ $HCl(g)$ $+$ $X(g)$
Where X=Noble gas
Hence the partical presence of $NH _3$ and $HCl$ will remain same.
The amount of gas in the system will increase after adding inert gas.
Since gases has high entropy than liquid and solid, the entropy  of the system will increase.  

Multiple choice

The equilibrium constant (K) for a chemical reaction is defined as:

  1. The ratio of products to reactants at equilibrium

  2. The ratio of reactants to products at equilibrium

  3. The difference between products and reactants at equilibrium

  4. The sum of products and reactants at equilibrium
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equilibrium constant (K) is defined as the ratio of products to reactants at equilibrium.

Multiple choice

The equilibrium constant for a chemical reaction is:

  1. The ratio of the concentrations of the reactants to the concentrations of the products.

  2. The ratio of the activities of the reactants to the activities of the products.

  3. The ratio of the partial pressures of the reactants to the partial pressures of the products.

  4. All of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equilibrium constant for a chemical reaction is the ratio of the concentrations, activities, or partial pressures of the reactants to the concentrations, activities, or partial pressures of the products.

Multiple choice

The equilibrium constant for a reaction is:

  1. The ratio of the concentrations of the reactants to the concentrations of the products.

  2. The ratio of the activities of the reactants to the activities of the products.

  3. The ratio of the partial pressures of the reactants to the partial pressures of the products.

  4. All of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equilibrium constant for a reaction is the ratio of the concentrations, activities, or partial pressures of the reactants to the concentrations, activities, or partial pressures of the products.

Multiple choice

What is the relationship between the equilibrium constant (K) and the concentrations of reactants and products at equilibrium?

  1. K = [products]/[reactants]

  2. K = [reactants]/[products]

  3. K = [products]^2/[reactants]^2

  4. K = [reactants]^2/[products]^2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equilibrium constant (K) is equal to the ratio of the concentrations of products and reactants at equilibrium.