Tag: introduction to equilibrium

Questions Related to introduction to equilibrium

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

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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

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

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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. $2$

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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

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Correct Option: B