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 evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

In the reaction, $N _2 + 3H _2 \rightarrow 2NH _3$, the ratio of volumes of nitrogen, hydrogen and ammonia is 1 : 3: 2. These figures illustrate the law of: 

  1. constant proportions

  2. Gay-Lussac

  3. multiple proportions

  4. reciprocal proportions
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

ANS; B
Gay Lussac’s Law of Combining Volumes states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the product(s) formed if gaseous, provided the temperature and pressure remain constant.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

Which of the following statements is/are correct?

  1. A sample of $CaCO _3$ contains $Ca = 40\%,\ C = 12\%$ and $O= 48\%$. If the law of constant composition is true, then the mass of $Ca$ in $10$ g of $CaCO _3$ from another source is $4.0$ g.
  2. $12$ g of carbon is heated in vacuum and there is no change in the mass. This is the best example of the law of conservation of mass.

  3. Air is heated at constant pressure and there is no change in mass but the volume increases. This is the best example of the law of conservation of mass.
  4. $SO _2$ gas was prepared by (i) heating $Cu $ with conc. $H _2SO _4$ (ii) burning sulphur in oxygen (iii) reacting sodium sulphite $(Na _2SO _3)$ with dilute $H _2SO _4$. It was observed that is each case, $S$ and $O$ combine in the ratio of $1:1$. This data illustrates the law of constant composition.
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

(A) A sample of $CaCO _3$ contains $Ca = 40\%$, $C = 12\%$ and $O= 48\%$. If the law of constant composition is true, then the mass of $Ca$ in $10$ g of $CaCO _3$ from another source is $\dfrac {40 \times 10 }{100}= 4.0 $ g


According to the law of constant composition, all samples of a given chemical compound have the same elemental composition by mass.
Hence, the statement A is true.

(B) $12$ g of carbon is heated in vacuum and there is no change in the mass. This is not the best example of the law of conservation of mass as there is no chemical transformation involved since there is a vacuum.
The Law of Conservation of Mass states that matter can be changed from one form into another, mixtures can be separated or made and pure substances can be decomposed but the total amount of mass remains constant.
Hence, the statement B is false.

(C) Air is heated at constant pressure and there is no change in mass but the volume increases. This is not the best example of the law of conservation of mass as there is no chemical transformation involved since there is a vacuum.

The Law of Conservation of Mass states that matter can be changed from one form into another, mixtures can be separated or made and pure substances can be decomposed but the total amount of mass remains constant.
Hence, the statement C is false.

(D) $SO _2$ gas was prepared by (i) heating $Cu$ with conc. $H _2SO _4$ (ii) burning sulphur in oxygen (iii) reacting sodium sulphite $(Na _2SO _3)$ with dilute $H _2SO _4$. It was observed that is each case, S and O combine in the ratio of $1:1$. This data illustrates the law of constant composition.

According to the law of constant composition, all samples of a given chemical compound have the same elemental composition by mass.

Hence, the statement D is true.

Multiple choice chemistry nitrogen and sulfur nitrogen gas component of air - nitrogen nitrogen

${ N } _{ 2 }(g)\ +\ { 3H } _{ 2 }(g)\rightleftharpoons { 2NH } _{ 3 }(g)$


For the reaction initially the mole was 1 : 3 of $N _2$ and ${ H } _{ 2 }$. At equilibrium 50% of each has reacted. If the equilibrium pressure is p, the partial pressure of ${ NH } _{ 3 }$ at equilibrium is :

  1. $\dfrac { p }{ 3 }$
  2. $\dfrac { p }{ 4 }$
  3. $\dfrac { p }{ 6 }$
  4. $\dfrac { p }{ 8 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$N _2 + 3H _2⇌2NH _3$
  1           3             0
(1-$x$)  ($3-3x$)     $2x$
            
$1-x =  \dfrac{1}{2}$
$\therefore x=\dfrac{1}{2}$
Mole of $H _2 = $ 1.5 mol
Mole of $N _2 = $ 0.5 mol
Mole of $NH _3 = $ 1.0 mol
Now,
Mole fraction of $NH _3$ =$\dfrac{1}{3}$

partial pressure = $p*\dfrac{1}{3}=\dfrac{p}{3}$
Multiple choice oxoacids of halogens p- block elements-ii p-block elements the p-block elements chemistry

${ COCl } _{ 2 }$ gas dissociates according to the equation
${ COCl } _{ 2 }(g)\ \rightleftharpoons \ CO(g)\ +\ { Cl } _{ 2 }(g)$
When ${ COCl } _{ 2 }$ is heated to 700 K at 100 kPa, density of the gas mixture at equilibrium is $1.2\ { gdm }^{ -3 }$. The percentage degree of dissociation of ${ COCl } _{ 2 }$ is: 

$(R = 8.314\ kPa\ { dm }^{ 3 } { K }^{ -1 }\ { mol }^{ -1 })$

  1. 41

  2. 81

  3. 60

  4. 20
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the ideal gas law PV=nRT and the relationship between density, molar mass, and degree of dissociation (alpha), one can solve for alpha. Given the density 1.2 g/dm3 at 700K and 100 kPa, the calculation yields approximately 41%.

Multiple choice chemistry chemical equilibrium and acids-bases dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

The degree of dissociation of $PCl _{5(g)}$ at 16.8 bar and $127^{0}C$ is 0.4. The value of $K _{P}$ for the reaction is:
$PCl _{5} \leftrightharpoons PCl _{3(g)} +Cl _{2(g)}$ 

  1. $3.2 bar$
  2. $3.2 bar^{-1}$
  3. $12.8\ bar$
  4. $ 0.4$ x $16.8\ bar$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

total pressure P total = $16.8 \, bar$

degree of dissociation $\alpha = 0.4$
$PCl _5 \rightleftharpoons p Cl _3 + Cl _2$
$P _0 (1 - \alpha) \,\,\, P _0 \alpha \,\,\, P _0 \alpha$
$P \, total = P _0(1 - \alpha) + P _0 \alpha + P _0 \alpha$
$= P _0 (1 + \alpha)$
$P _0 (1 + \alpha) = 16.8$
$P _0 \times 1.4 = 16.8 \Rightarrow P _0 = \dfrac{16.8}{1.4} = 12 $ bar
$Kp = \dfrac{[PCl _3][Cl _2]}{[PCl _5]} = \dfrac{P _0 \alpha \times P _0 \alpha}{P _0 (1 - \alpha)}$
$= \dfrac{P _0 \alpha^2}{1 - \alpha}$
$= 12 \times \dfrac{0.4 \times 0.4}{0.6}$
$= 3.2 \, bar$

Multiple choice chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

Consider the reaction $2A+B$ $\rightarrow$products,when the concentration of a alone was doubled, the half-life of the  reaction did not change.When the concentration of B alone was double,the rate was not altered.The unit of rate constant for this reaction is

  1. $S^{-1}$
  2. $L\ mol^{-1}\ s^{-1}$
  3. $mol\ L^{-1}\ s^{-1}$
  4. $mol^{-2}\ L^{5}\ S^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

The following two equilibria exist simultaneously in a closed vessel :
$PCI _5(g) \rightleftharpoons PCI _3(g) + Cl _2(g)$
$COCI _2(g) \rightleftharpoons CO (g) + CI _2 (g)$
If some CO is added into the vessel, then after the equlibrium is attained again, concertration of ?

  1. $PCI _5$ will increase
  2. $PCI _5$ will decrease
  3. $PCI _5$ will remain unaffected
  4. $CI _2$ will increase
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given reactions are,

$PCl _5 \rightleftharpoons PCl _3+Cl _2$
$COCl _2\rightleftharpoons CO _{(g)}+Cl _2$
If some $CO$ is added into the vessel, the concentration of $PCl _5$ will remain unaffected, because $CO$ reacts with $Cl _2$ and forms $COCl _2$, not going to effect the concentration of $PCl _5$ .

Multiple choice chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

In a closed system : $A\left( s \right) \rightleftharpoons 2B\left( g \right) +3C\left( g \right) $ if the partial pressure C is of doubled then partial pressure B wil be:

  1. Twice the original pressure

  2. Half of its original pressure
  3. $\dfrac { 1 }{ 2\sqrt { 2 } } $ times, the original pressure
  4. $2\sqrt { 2 } $ times its original pressure
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Solution:- (C) $\cfrac{1}{2 \sqrt{2}}$ times, the original pressure
${A} _{\left( s \right)} \rightleftharpoons 2 {B} _{\left( g \right)} + 3 {C} _{\left( g \right)}$
${K} _{P} = {\left( {P} _{B} \right)}^{2} {\left( {P} _{C} \right)}^{3} ..... \left( 1 \right)$
If we double the partial pressure of $C$, i.e., ${P} _{C}' = 2 {P} _{C}$
$\therefore {K} _{P}' = {\left( {P} _{B}' \right)}^{2} {\left( {P} _{C}' \right)}^{3}$
$\Rightarrow {K} _{P}' = {\left( {P} _{B}' \right)}^{2} {\left( 2 {P} _{C} \right)}^{3}$
$\Rightarrow {K} _{P}' = 8 {\left( {P} _{B}' \right)}^{2} {\left( {P} _{C} \right)}^{3}$
Since ${K} _{P}$ is constant,
$\therefore {K} _{P} = {K} _{P}'$
$\Rightarrow {\left( {P} _{B} \right)}^{2} {\left( {P} _{C} \right)}^{3} = 8 {\left( {P} _{B}' \right)}^{2} {\left( {P} _{C} \right)}^{3}$
$\Rightarrow {P} _{B}' = \sqrt{\cfrac{{P} _{B}}{8}}$
$\Rightarrow {P} _{B}' = \cfrac{{P} _{B}}{2 \sqrt{2}}$
Hence the partial pressure of $B$ will be $\cfrac{1}{2 \sqrt{2}}$ times of its original pressure.
Multiple choice chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

Ammonium carbamate dissociates as ${ NH } _{ 2 }COON{ H } _{ 4\left( s \right)  }\leftrightharpoons 2N{ H } _{ 3\left( g \right)  }+{ CO } _{ 2\left( g \right)  }$. In a closed vessel containing ammonium carbamate in equilibrium, ammonia is added such that the partial pressure of ${ NH } _{ 3 }$ now equals to the original total pressure. The ratio of total pressure now to the original pressure is :

  1. $\frac { 27 }{ 31 } $
  2. $\frac { 31 }{ 27 } $
  3. $\frac { 4 }{ 9 } $
  4. $\frac { 5 }{ 9 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the reaction NH2COONH4(s) <=> 2NH3(g) + CO2(g), the initial equilibrium partial pressures are P(NH3) = 2p and P(CO2) = p, so total pressure P1 = 3p. After adding NH3, the new P(NH3) = 3p. Using Kp = (2p)^2 * p = 4p^3, the new equilibrium satisfies 4p^3 = (3p)^2 * P(CO2_new), giving P(CO2_new) = 4p/9. The new total pressure P2 = 3p + 4p/9 = 31p/9. The ratio P2/P1 = (31p/9) / 3p = 31/27.

Multiple choice chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

$5$ moles of $SO _{2}$ and 5 moles of $O _{2}$ are allowed to react to form $SO _{3}$  in a closed vessel. At the equilibrium stage, $60\%$ $SO _{2}$ is used up. The total number of moles of $SO _{2}$, $O _{2}$ and $SO _{3}$ in the vessel now is

  1. $10.5$
  2. $10.0$
  3. $8.5$
  4. $3.9$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Initially, we have 5 moles of SO2 and 5 moles of O2. Since 60 percent of SO2 is used up, 0.60 times 5 equals 3 moles of SO2 reacted. Using stoichiometry (2 SO2 + O2 -> 2 SO3), 3 moles of SO2 require 1.5 moles of O2 and produce 3 moles of SO3. The remaining moles are: SO2 = 5 - 3 = 2, O2 = 5 - 1.5 = 3.5, and SO3 = 3, giving a total of 2 + 3.5 + 3 = 8.5 moles.

Multiple choice real gases van der-waal equation: equation of state for real gas kinetic theory of gases thermal physics physics

For gaseous decomposition of ${PCI} _{5}$ in a closed vessel the degree of dissociation '$\alpha $', equilibrium pressure 'P' & ${'K} _{p}'$ are related as

  1. $\\ \alpha =\sqrt { \frac { { K } _{ p } }{ P } } $
  2. $\\ \alpha =\frac { 1 }{ \sqrt { { K } _{ p }+P } } $
  3. $\\ \alpha =\sqrt { \frac { { K } _{ p }+P }{ { K } _{ p } } } $
  4. $\alpha =\sqrt { { K } _{ p }+P } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${  \quad \quad \quad \quad \quad \quad \quad PCl } _{ 5 }\rightleftharpoons { PCl } _{ 3(9) }+{ Cl } _{ 2(9) }\\ Initial\quad mole\quad \quad \quad 1\quad  \quad 0\quad\quad\quad 0\\ After\quad mole\quad \quad 1-\alpha \quad \quad  \alpha \quad\quad  \alpha \\ decomposition$

Total mole$=1-\alpha+\alpha+\alpha\\=1+\alpha$

Total pressure$=P$

Partial pressure of $PCl _5=P(\cfrac{1-\alpha}{1+\alpha})$

PArtial pressure of $PCl _3=P(\cfrac{\alpha}{1+\alpha})$

Partial pressure of $PCl _2=P(\cfrac{\alpha}{1+\alpha})$

Then $K _p=\cfrac{(PCl _3)(Cl _2)}{(PCl _5)}\\ \quad=\cfrac{P(\cfrac{\alpha}{1+\alpha})P(\cfrac{\alpha}{1+\alpha})}{P(\cfrac{1-\alpha}{1+\alpha})}\\ \quad=\cfrac{P^2\alpha^2}{(1+\alpha)^2}\times\cfrac{(1+\alpha)}{P(1-\alpha)}\\K _p=\cfrac{P\alpha^2}{1-\alpha^2}$

now, $1-\alpha^2<<1$

so that $K _p=P\alpha^2\\ \alpha^2=\cfrac{K _p}{P}\\ \alpha=\sqrt{\cfrac{K _p}{P}}$

 

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.