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 chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

If two gases $AB _2$ and $B _2C$ are mixed the following equilibria are readily established
$AB _2(g) + B _2 C(g)  \rightarrow AB _3(g) + BC(g)$
$BC(g) + B _2 C(g)  \rightarrow B _3 C _2 (g)$
If the reaction is started only with $AB _2$ with $B _2C$, then which of the following is necessarily true at equilibrium:

  1. $[AB _3] _{eq} = [BC] _{eq}$
  2. $[AB _2] _{eq} = [B _2C] _{eq}$
  3. $[AB _3] _{eq} > [B _3C _2] _{eq}$
  4. $[AB _3] _{eq} > [BC] _{eq}$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

Let reactions is started with a mole of $AB _2$ and b mole of $B _2C$
$\Rightarrow      AB _2 (g) + B _2C(g)  \rightarrow AB _3(g) + BC(g)$
                    a                 b               0               0
                  a - x           b - x - y        x              x - y
$BC(g) + B _2C(g)   \rightarrow B _2C _2 (g)$
    x - y            b - x - y        y                 As  x > y
Clearly $[AB _3] _{eq} > [B _3 C _2] _{eq} $ and  $[AB _3] _{eq}  >  [BC] _{eq}$

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

The equilibrium constant for a reaction is $K$, and the reaction quotient is $Q$. For a reaction mixture, the ratio $\dfrac {K}{Q}$ is $0.33$. This means that:

  1. the reaction mixture will equilibrium to form more reactant species

  2. the reaction mixture will equilibrium to form more product species
  3. the equilibrium ratio of reactant to product concentrations will be $3$
  4. the equilibrium ratio of reactant to product concentrations will be $0.33$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that for a reaction if $\dfrac{K}{Q} < 1,$ the reaction proceeds in backward direction. Hence, the reaction mixture forms more reactant species.

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

Assume that the decomposition of $H{ NO } _{ 3 }$ can be represented by the following equation
$4H{ NO } _{ 3 }(g)\rightleftharpoons 4{ NO } _{ 2 }(g)+2{ H } _{ 2 }O(g)+{ O } _{ 2 }(g)\quad $'and the reaction approaches equilibrium at $400K$ temperature and $30$ atm pressure. The equilibrium partial pressure of $H{ NO } _{ 3 }$ is $2$ atm
Calculate ${K} _{c}$ in ${ \left( mol/L \right)  }^{ 3 }$
(Use: $R=0.08atm-L/mol-K$)

  1. $4$
  2. $8$
  3. $16$
  4. $32$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

The optical rotation of the $\alpha-form$ of a pyramose is $+150.7^{\circ}$, that of the $\beta - form$ is $+52.8^{\circ}$. In solution an equilibrium mixture of these anomers has an optical rotation of $+80.2^{\circ}$. The percentage of the $\alpha$ form in equilibrium mixture is:

  1. $28$%
  2. $32$%
  3. $68$%
  4. $72$%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\alpha $ from $=+150.{ 7 }^{ 0 }$, $\beta $ from $=+52.{ 8 }^{ 0 }$

at equilibrium optical rotation $=+80.2$ 
let, at equilibrium $\alpha $-from exist $=x$
      at equilibrium $\beta $-from exist $=(100-x)$
Therefore, $\dfrac { 150.7x+\left( 100-x \right) \times 52.8 }{ 100 } =80.2$
$\Rightarrow \quad 150.7x+5280-52.8x=8020$
$\Rightarrow \quad 99.9x=2740$
$\Rightarrow \quad x=27.42\approx 28$%

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

The equilibrium constant $K _{c}$ for the reaction $P _{4}(g) \rightleftharpoons 2P _{2}(g)$
is $1.4$ at $400^{\circ}C$. Suppose that $3$ moles of $P _{4}(g)$ and $2$ moles of $P _{2}(g)$ are mixed in $2$ litre container at $400^{\circ}C$. What is the value of reaction quotient $(Q _{c})$?

  1. $\dfrac {3}{2}$
  2. $\dfrac {2}{3}$
  3. $1$

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

$Q _c = \dfrac{[P _2(g)]^4}{[P _4 (g)]}$

= $\dfrac{(1)^2}{(3/2)}$ 
= $\dfrac{2}{3}$

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

0.1 mole of $N _2O _4(g)$ was sealed in a tube under one atmospheric conditions at $25^0C$. Calculate the number of moles of $NO _2(g)$ present, if the equilibrium $N _2O _4(g)\rightleftharpoons  2NO _2(g)$ $(K _p=0.14)$ is reached after some time.

  1. $1.8 \times 10^2$
  2. $2.8 \times 10^2$

  3. 0.034
  4. $2.8 \times 10^{-2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to the equation

$\begin{matrix}  & N _2O _4& \rightleftharpoons & 2NO _2 \ \text{Initial}&1 \, atm  & &0 \ \text{change}& -x&  &+2x\ \text{Equilibrium}&1-x&&2x  \end{matrix}$
$K _P = \dfrac{(P _{NO _2})^2}{P _{N _2O _4}}$   $[\therefore K _P$ is similar to $K _C$ in aspect of setting up rate quotient]

$0.14 = \dfrac{(2x)^2}{1-x}$

$0.14(1-x) = 4x^2$
$x = 0.17$
From ideal gas equation
$PV = nRT$

$V = \dfrac{nRT}{P}$

$V = \dfrac{0.1\times 0.082 \times (273 + 25)}{1}$

$V = 2.45 lit$

Moiles of $NO _2= \dfrac{P _{NO _2} \times V}{RT}$

Moles of $NO _2 = \dfrac{0.34\times 2.45}{0.0821 \times (273 + 25)}$

Moles of $NO _2 = 0.034$

$\therefore$ option C is correct

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

Consider the following reactions in which all the reactants and the products are in 


$2PQ \rightleftharpoons P _2 + Q _2 ; K _1 = 2.5 \times 10^5$

$PQ + \cfrac{1}{2} R _2 \rightleftharpoons PQR; K _2 = 5 \times 10^{-3}$

The value of $K _3$ for the equilibrium $\cfrac{1}{2} P _2 + \cfrac{1}{2} Q _2 + \cfrac{1}{2} R _2 \rightleftharpoons PQR,$ is:

  1. $2.5 \times 10^{-3}$
  2. $2.5 \times 10^{3}$
  3. $1.0 \times 10^{-5}$
  4. $5 \times 10^{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

Which of the following statements is correct?

  1. In equilibrium mixture of ice and water kept in perfectly insulated flask, mass of ice and water does not change with time.

  2. The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate.

  3. On addition of catalyst, the equilibrium constant value is not affected.
  4. Equilibrium constant for a reaction with negative $\triangle$H value decreases as the temperature increases.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equilibrium reaction containing $Fe(III)$ nitrate & $KSCN$ is given as :-

${ Fe }^{ 3+ }\left( aq \right) +{ SCN }^{ - }\left( aq \right) \rightleftharpoons { \left[ Fe\left( SCN \right)  \right]  }^{ 2+ }\left( aq \right) $
Here, the red colour in the reactions occurs due to formation of ${ \left[ Fe\left( SCN \right)  \right]  }^{ 2+ }$. Now, when oxalic acid is added, it will react with ${ Fe }^{ 3+ }$ to form ${ \left[ Fe{ \left( { C } _{ 2 }{ O } _{ 4 } \right)  } _{ 3 } \right]  }^{ 3- }$. So, the concentration of ${ Fe }^{ 3+ }$ ions will get decreased. By applying Lechatelier's principle we see that the reaction will proceed towards backward direction and concentration of ${ \left[ Fe\left( SCN \right)  \right]  }^{ 2+ }$ will get decreased. Consequently the intensity of red colour is decreased.

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

When sulphur ( in the form of $S _8$) is heated to temperature T, at equilibrium, the pressure of $S _8$ falls by 30% from 1.0 atm, because $S _8$(g) is partially converted into $S _2$(g). Find the value of $K _p$ for this reaction.

  1. $2.96$
  2. $6.14$
  3. $204.8$

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

                                                  ${{S} _{8}}\rightleftharpoons 4{{S} _{2}}$

Initial pressure                          $1$atm     $0$

At equilibrium pressure    $(1-0.3)$     $4\times 0.3$


Therefore, equilibrium pressure of ${{S} _{8}}=(1-0.3)=0.7$ atm

And equilibrium pressure of ${{S} _{2}}=4\times 0.3=1.2$ atm


So, equilibrium constant $Kp=\dfrac{P _{S _2}^4}{P _{S _8}}=\dfrac{{{1.2}^{4}}}{0.7}=$ approximate $2.96$ atm$^3$. 

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

At $298K$, the equilibrium constant of reaction.
${ Zn }^{ +2 }+4{ NH } _{ 3 }\rightleftharpoons { \left[ Zn{ \left( { NH } _{ 3 } \right)  } _{ 4 } \right]  }^{ +2 }$ is ${ 10 }^{ 9 }$
If ${ E } _{ { \left[ Zn{ \left( { NH } _{ 3 } \right)  } _{ 4 } \right]  }^{ +2 }/Zn+4{ NH } _{ 3 } }^{ o }=-1.03V$. The value ${ E } _{ Zn/{ Zn }^{ +2 } }$ will be: 

  1. 0.7645V

  2. -1.1V

  3. +1.1V

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

The reaction is,

$Zn\rightarrow { Zn }^{ 2+ }+2e$
Since, given ${ { E }^{ 0 } } _{ { \left[ Zn{ \left( { NH } _{ 3 } \right)  } _{ 4 } \right]  }^{ 2+ }/Zn+4{ NH } _{ 3 } }=-1.03V$
${ K } _{ eq }={ 10 }^{ 9 }$
We know,
$E={ E }^{ 0 }+\dfrac { 0.059 }{ n } log{ K } _{ eq }$
$E=-1.03+\dfrac { 0.059 }{ 2 } \times 9$         since, the reaction is occured transfaring two electron.
$E=-0.7645V$

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

${ K } _{ c }$ for the reaction $A+B\overset { { K } _{ 1 } }{ \underset { { K } _{ 2 } }{ \rightleftharpoons  }  }  C+D$ , is equal to: 

  1. $\dfrac {{ K } _{ 1 }}{ { K } _{ 2 }}$
  2. $K _{ 1 }{ K } _{ 2 }$
  3. $K _{ 1 }-{ K } _{ 2 }$
  4. $K _{ 1 }+{ K } _{ 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
${ K } _{ C }=$ Equilibrium constant
$A+B\overset { { K } _{ 1 } }{ \underset { { K } _{ 2 } }{ \rightleftharpoons  }  } \quad C+D$
${ K } _{ C }=\cfrac { { K } _{ 1 } }{ { K } _{ 2 } } =\cfrac { \left[ C \right] \left[ D \right]  }{ \left[ A \right] \left[ B \right]  } $
As at equilibrium,
Rate of forward reaction=rate of backward reaction
${ r } _{ f }={ r } _{ b }$
${ K } _{ 1 }\left[ A \right] \left[ B \right] ={ K } _{ 2 }\left[ C \right] \left[ D \right] $
${ K } _{ C }=\cfrac { { K } _{ 1 } }{ { K } _{ 2 } } =\cfrac { \left[ A \right] \left[ B \right]  }{ \left[ C \right] \left[ D \right]  } $
There, option $A$ is correct.
Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

$PCl _5(g)\rightleftharpoons PCl _3(g)\,+\,Cl _2(g)$

In the above reaction taking place in a closed rigid vessel, at constant temperature, starting with $PCl _5$ initially, which of the following is correct observations with the progress of reaction?

  1. Average molar mass increases

  2. Total number of moles increases

  3. Pressure remains constant
  4. Partial pressure of $PCl _5$ increases and that of $PCl _3$ decreases
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In PCl5 -> PCl3 + Cl2, one mole of gas produces two moles of gas. Thus, the total number of moles increases.

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

$3C _2H _2\rightleftharpoons C _6H _6$ 


The above reaction is performed in a 1-liter vessel. Equilibrium is established when $0.5\ mole$ of benzene is present at a certain temperature. If the equilibrium constant is $4\ L^2mol^{-2}$. The total number of mole of the substance present at equilibrium is:

  1. $0.5$
  2. $1$
  3. $1.5$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

Which of the following reactions have same heat of reaction at constant $P$ and constant volume as well?

  1. $2NO(g)\longrightarrow N _2(g)+O _2(g)$
  2. $N _2(g)+3H _2(g)\longrightarrow 2NH _3(g)$
  3. $Co _3O _4(s)+4CO(g)\longrightarrow 3Co(s)+4CO _2(g)$
  4. $H _2(g)+Cl _2(g)\longrightarrow 2HCl(g)$
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

The following reactions have the same heat of reaction at constant $P$ and constant volume as well.


$2NO(g)\longrightarrow N _2(g)+O _2(g)$
$Co _3O _4(s)+4CO(g)\longrightarrow 3Co(s)+4CO _2(g)$
$H _2(g)+Cl _2(g)\longrightarrow 2HCl(g)$

This is because, in these reactions, the number of moles of gaseous reactants and the number of moles of gaseous products is the same.

$\because \Delta n _g = 0$

However, for the reaction $N _2(g)+3H _2(g)\longrightarrow 2NH _3(g)$, heat of reaction at constant $P$ and constant volume are different.

This is because, in these reactions, the number of moles of gaseous reactants and the number of moles of gaseous products are different.