Tag: chemistry

Questions Related to chemistry

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.

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

  1. $1.8 \times 10^2$

  2. $2.8 \times 10^2$

  3. 0.034

  4. $2.8 \times 10^{-2}$

Show answer
Correct Option: C
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

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:

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

  1. $2.5 \times 10^{-3}$

  2. $2.5 \times 10^{3}$

  3. $1.0 \times 10^{-5}$

  4. $5 \times 10^{3}$

Show answer
Correct Option: A

Chemical equilibria is (I) _________ in nature. It occurs in (2) ________ reactions only.  At equilibria, all (3) __________ properties becomes constant. Chemical equilibrium gets affected by change in temperature, pressure, concentration, volume etc but is not altered by addition of (4)________. Equilibria are of two types (5) ________ and homogeneous equilibria.

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

  1. (1) dynamic (2) reversible (3) observable (4) catalyst (5) homogeneous 

  2. (1) static (2) irreversible (3) observable (4) reactants (5) heterogeneous

  3. (1) dynamic (2) irreversible (3) observable (4) reactants (5) homogeneous 

  4. (1) dynamic (2) reversible (3) observable (4) catalyst (5) heterogeneous 

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Correct Option: D
Explanation:

Chemical equilibria is dynamic in nature. It occurs in reversible reactions only. At equilibria all observable properties becomes constant. Chemical equilibrium is affected by a change in temperature, pressure, concentration, volume etc but is not altered by the addition of catalyst. Equilibria is of two types heterogeneous and homogeneous equilibria.

Which of the following statements is correct?

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

  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.

Show answer
Correct Option: B
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.

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

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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