Tag: introduction to equilibrium

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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.