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 the s-block elements (alkali and alkaline earth metals) some important compounds of calcium some important compounds of magnesium and calcium compounds of s block elements

$A \, Ca(HCO _3) _2 + B \, Ca(OH) _2 \to C \, CaCO _3 + D \, H _2O$. 


Choose the correct option.

  1. $A = 1$
  2. $B = 1$
  3. $C = 1$
  4. $D = 2$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

$ACa(HCO _3) _2 + BCa(OH) _2 \rightarrow CCaCO _3 + DH _2O$

First, balance calcium then oxygen ,then hydrogen.

$Ca(HCO _3) _2 + Ca(OH) _2 \rightarrow 2CaCO _3 + 2H _2O$
                       (This the balanced equation)

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

In Kjeldahl's method, nitrogen present is estimated as:

  1. $N _2$
  2. $NH _2$
  3. $NO _2$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Kjeldahl method is a method that is used the quantitatuin estimation of nitrogen conlaining in inorganic substances like $NH _3/NH _4^+$, and in organic substances.In kjeldah's method, notrogen is estimated in following steps : 

$Step$ $1:$Degradation: Sample$+H _SO _4\longrightarrow {(NH _4)} _2SO _4+CO _2+SO _2+H _2O$
$Step$ $2:$ Ammonia is liberated : ${(NH _4)} _2SO _4+2NaOH\longrightarrow Na _2SO _4+2H _2O+2NH _3$
$Step$ $3:$ Ammonia is captured : $B{(OH)} _3+H _2O+NH _3\longrightarrow NH^+ _4+B{(OH)} _4^-$
$Step$ $4:$ Back titration : $B{(OH)} _3+H _2O+Na _2CO _3\longrightarrow NaHCO _3+NaB{(OH)} _4+CO _2+H _2O$
Thus in Kjeldahl's method, nitrogen present is estimated as $NH _2$.

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

For a particular $A+B \rightarrow C$ was studied at $25^{\circ}C$. The following results are obtained.


              [A]              [B]           [C]
    (mole/lit)       (moles/lit)  (mole  lit $^{-1} sec^{-2}$)  
$9 \times 10^{-5}$ $1.5 \times 10^{-2}$           $0.06$
$9 \times 10^{-5}$ $3 \times 10^{-3}$            $0.012$
$3 \times 10^{-5}$ $3 \times 10^{-3}$            $0.004$
$6 \times 10^{-5}$            x           $0.024$


Then the value of x is :

  1. $6 \times 10^{-3} moles litre^{-1}$
  2. $3 \times 10^{-3} moleslitre^{-1}$
  3. $4.5 \times 10^{-3} moleslitre^{-1}$
  4. $9 \times 10^{-3} moleslitre^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$A+B\rightarrow C$

$ rate=k\left[ A \right] \left[ B \right] $

$Experiment \  3\& 2 \  chosen \  for \  value \  of \  k \  as\left[ B \right] is \  same \  in \  both$ 

$\dfrac { { r } _{ 3 } }{ { r } _{ 2 } } =\dfrac { 0.004 }{ 0.012 } =k\dfrac { \left[ { 3\times 10 }^{ -5 } \right] \left[ { 3\times 10 }^{ -3 } \right]  }{ \left[ { 9\times 10 }^{ -5 } \right] \left[ { 3\times 10 }^{ -3 } \right]  } $

$k=1 \ using \  this \  rate \  constant \  value \  in \  finding \  x\\$
$ \dfrac { { r } _{ 4 } }{ { r } _{ 3 } } =\dfrac { 0.024 }{ 0.004 } =k\dfrac { \left[ { 6\times 10 }^{ -5 } \right] \left[ x \right]  }{ \left[ { 3\times 10 }^{ -5 } \right] \left[ { 3\times 10 }^{ -3 } \right]  } $

$\\ \left[ x \right] ={ 9\times 10 }^{ -3 }\\ $
Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

$2K(s)+2{H} _{2}O(l)\rightarrow 2KOH(aq)+{H} _{2}(g)$
If $3.0$ moles of potassium react with excess water, what volume of hydrogen gas will be produced?

  1. $1.5L$
  2. $22.4L$
  3. $67.2L$
  4. $33.6L$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given, $2K(s) + 2H _2O(l) \rightarrow 2KOH(aq) + H _2 (g)$

$2\space moles$ of Potassium on reaction gives $1 \space mole$ of Hydrogen [ 22.4 at STP]
$2\space moles$ of $K \rightarrow 22.4 \space Litres $ of $H _2$
$2\space moles$ of $K = \dfrac{3\times 22.4}{2} = 33.6 \space Litres$ of $H _2$
So, $33.6\space Litres$ of Hydrogen are produced.

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