Tag: chemical equilibrium and acids-bases

Quizzes (7) Questions (149)

Questions Related to chemical equilibrium and acids-bases

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

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

  1. differs with substance

  2. changes

  3. constant

  4. equilibrium
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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: 

  1. $8 \times 10^{-5} atm$
  2. $11.3 \times 10^{-7} atm$
  3. $9.71 \times 10^{-6} atm$
  4. $8 \times 10^{-2} atm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Kp = P_O2^3 / P_O3^2 = 4 * 10^14. Total pressure = P_O2 + P_O3 = 8. Since P_O2 >> P_O3, P_O2 approx 8. 8^3 / P_O3^2 = 4 * 10^14. 512 / P_O3^2 = 4 * 10^14. P_O3^2 = 128 * 10^-14. P_O3 = sqrt(128) * 10^-7 = 11.3 * 10^-7.

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
B Correct answer
Explanation

Given the reaction 3 C2H2 <=> C6H6 with Kc = 4 and 0.5 moles of benzene at equilibrium in a 1-liter vessel. Setting up an ICE table, if benzene is 0.5, acetylene reacted is 1.5, leaving 0 for acetylene at equilibrium, leading to a total of 0.5 + 0.5 = 1 mole if calculated correctly with initial values, or simply using stoichiometry and equilibrium concentrations.

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

${ H } _{ 3 }{ PO } _{ 4 }$ 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 $P=31$)

  1. $100\ ml$
  2. $200\ ml$
  3. $80\ ml$
  4. $300\ ml$
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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 $Na _{2}CrO _{4}$?

  1. $Ag _{2}CrO _{4}$
  2. $AgI$
  3. $AgCl$
  4. $AgBr$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.