Tag: chemical equilibrium and acids-bases

Questions Related to 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.

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

  1. $2.96$

  2. $6.14$

  3. $204.8$

  4. None of these

Show answer
Correct Option: A
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$^3$. 

In a dynamic equilibrium, the concentrations of reactants and products remains ___________.

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

  1. differs with substance

  2. changes

  3. constant

  4. equilibrium

Show answer
Correct Option: C
Explanation:

In chemistry, a dynamic equilibrium exists once a reversible reaction ceases to change its ratio of reactants/products, but substances move between the chemicals at an equal rate, meaning there is no net change. It is a particular example of a system in a steady state. In thermodynamics a closed system is in thermodynamic equilibrium when reactions occur at such rates that the composition of the mixture does not change with time. Reactions do in fact occur, sometimes vigorously, but to such an extent that changes in composition cannot be observed.

Hence, in a dynamic equilibrium, the concentrations of reactants and products remains constant.

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: 

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

  1. 0.7645V

  2. -1.1V

  3. +1.1V

  4. none of these

Show answer
Correct Option: D
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.
$E=-0.7645V$

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

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

  1. $\dfrac {{ K } _{ 1 }}{ { K } _{ 2 }}$

  2. $K _{ 1 }{ K } _{ 2 }$

  3. $K _{ 1 }-{ K } _{ 2 }$

  4. $K _{ 1 }+{ K } _{ 2 }$

Show answer
Correct Option: A
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.

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

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?

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

  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

Show answer
Correct Option: B

A $10\ litre$ box contains $O _3$ and $O _2$ at equilibrium at 2000 K. $K _p=4 \times 10^{14}$ atm for $2O _3(g) \rightleftharpoons  3O _2(g)$. Assume that $P _{O _2} > > P _{O _3}$ and if total pressure is 8 atm, then patial pressure of $O _3$ will be: 

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

  1. $8 \times 10^{-5} atm$

  2. $11.3 \times 10^{-7} atm$

  3. $9.71 \times 10^{-6} atm$

  4. $8 \times 10^{-2} atm$

Show answer
Correct Option: B

$3C _2H _2\rightleftharpoons C _6H _6$ 


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:

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

  1. $0.5$

  2. $1$

  3. $1.5$

  4. $2$

Show answer
Correct Option: A

In $ A _{3}(g) \leftrightharpoons3A(g) $ reaction, the initial concentration of $ A _{3} $ is "a" and $ molL^-1$ If x is degree of dissociation of $ A _{3} $. The total number of moles at equilibrium will be:- 

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

  1. $a-\frac{ax}{3}$

  2. $\frac{2ax}{3}-3$

  3. $\frac{a-ax}{2}$

  4. None of these

Show answer
Correct Option: B

${ H } _{ 3 }{ PO } _{ 4 }$ is a tribasic acid and one of its salt is ${ NaH } _{ 2 }{ PO } _{ 4 }$ What volume of $1M\quad NaOH$ solution should be added to $12g\ { NaH } _{ 2 }{ PO } _{ 4 }$ to convert it into ${ Na } _{ 3 }{ PO } _{ 4 }$? ($at.wt$ of $P=31$)

chemistry water drying and dehydrating agents hydrolysis of salts and the ph of their solutions chemical equilibrium and acids-bases

  1. $100\ ml$

  2. $200\ ml$

  3. $80\ ml$

  4. $300\ ml$

Show answer
Correct Option: B
Explanation:

$Na{ H } _{ 2 }{ PO } _{ 4 }+NaOH\rightarrow { Na } _{ 2 }{ HPO } _{ 4 }+{ H } _{ 2 }O\quad \quad -(i)$

Moles$=\cfrac { 12 }{ 120 } $    $\cfrac { 12 }{ 120 } $            $\cfrac { 12 }{ 120 } $
${ Na } _{ 2 }{ HPO } _{ 4 }+NaOH\rightarrow { Na } _{ 3 }{ PO } _{ 4 }+{ H } _{ 2 }O\quad \quad \quad -(ii)$
Moles$=\cfrac { 12 }{ 120 } $     $\cfrac { 12 }{ 120 } $           $\cfrac { 12 }{ 120 } $
Moles of $NaOH$ required $=\cfrac { 12 }{ 120 } +\cfrac { 12 }{ 120 } $
                                              $=0.2$ moles
Now, $Mularity=\cfrac { Moles }{ Vol.of\quad sol\quad in\quad ltrs } $
$\therefore $ Volume of $NaOH$ solution $=\cfrac { 0.2 }{ 1 } $
                                                  $=0.2$ litres
                                                  $=200$ml

The $K _{sp}$ of $Ag _{2}CrO _{4}, AgCl, AgBr$ and $AgI$ are respectively, $1.1\times 10^{-12}$, $1.8\times 10^{-10}$, $5.0\times 10^{-13}$ and $8.3\times 10^{-17}$. Which of the following salts will precipitate last if $AgNO _{3}$ solution is added to the solution containing equal moles of $NaCl, NaBr, NaI$ and $Na _{2}CrO _{4}$?

chemistry water drying and dehydrating agents hydrolysis of salts and the ph of their solutions chemical equilibrium and acids-bases

  1. $Ag _{2}CrO _{4}$

  2. $AgI$

  3. $AgCl$

  4. $AgBr$

Show answer
Correct Option: A
Explanation:
  1. $Ag _2CrO _4\rightleftharpoons 2Ag^++{CrO _4}^{2-}$

    $Ksp=(2s)^2\times s=4s^2$

    $Ksp=(1.1\times 10^{-12})$

    $S=3\sqrt {\cfrac {Ksp}{4}}=6.5\times 10^{-5}$M

    2. $AgCl\rightleftharpoons Ag^++Cl^-$

    $Ksp=S\times S$  ;       $Ksp=1.8\times 10^{-10}$

    $S=\sqrt {Ksp}=1.34\times 10^{-5}$M

    3. $AgBr\rightleftharpoons Ag^++Br^-$

    $Ksp=S\times S$ ;     $Ksp=5\times 10^{-13}$

    $S=\sqrt {Ksp}=0.71 \times 10^{-6}$M

    4. $AgI\rightleftharpoons Ag^++I^-$

    $Ksp=S\times S$  ;    $Ksp=8.3\times 10^{-17}$

    $S=\sqrt{Ksp}=0.9\times 10^{-8}$M

    $\therefore$ Solubility of $Ag _2CrO _4$ is highest, so it will precipitate last.