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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
C Correct answer
Explanation

To obtain the target equilibrium equation, multiply the first reaction by -1/2 (or reverse and take square root) and add the second reaction. This corresponds to combining the equilibrium constants as K3 = (1 / sqrt(K1)) * K2.

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

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.

  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

Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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.