Chemistry

Chemical Equilibrium and Stoichiometry

103 Questions

Chemical equilibrium and stoichiometry questions cover equilibrium constants, percent yield, and the Haber process. These concepts test your ability to calculate reaction outcomes and purity. Practice these to excel in physical chemistry sections of competitive tests.

Equilibrium constant calculationsStoichiometric yieldHaber processPercent purityHomogenous equilibrium

Chemical Equilibrium and Stoichiometry Questions

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

$3C _2H _2\rightleftharpoons C _6H _6$ 


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 enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

Which of the following reactions have same heat of reaction at constant $P$ and constant volume as well?

  1. $2NO(g)\longrightarrow N _2(g)+O _2(g)$
  2. $N _2(g)+3H _2(g)\longrightarrow 2NH _3(g)$
  3. $Co _3O _4(s)+4CO(g)\longrightarrow 3Co(s)+4CO _2(g)$
  4. $H _2(g)+Cl _2(g)\longrightarrow 2HCl(g)$
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

The following reactions have the same heat of reaction at constant $P$ and constant volume as well.


$2NO(g)\longrightarrow N _2(g)+O _2(g)$
$Co _3O _4(s)+4CO(g)\longrightarrow 3Co(s)+4CO _2(g)$
$H _2(g)+Cl _2(g)\longrightarrow 2HCl(g)$

This is because, in these reactions, the number of moles of gaseous reactants and the number of moles of gaseous products is the same.

$\because \Delta n _g = 0$

However, for the reaction $N _2(g)+3H _2(g)\longrightarrow 2NH _3(g)$, heat of reaction at constant $P$ and constant volume are different.

This is because, in these reactions, the number of moles of gaseous reactants and the number of moles of gaseous products are different.

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

Pure $PCl _5(g)$ was placed in a flask at $2\,atm$ and left for some time at a cretain temp. Where the following equilibrium was established $PCl _5(g) \rightleftharpoons PCl _3(g)+Cl _2(g)$
Under the given condition,pure oxygen was found to effuse $2.084$ time faster than the equilibrium mixture then $K _p$ of the above reaction is:

  1. $\dfrac{2}{3}$
  2. $\dfrac{3}{2}$
  3. $\dfrac{4}{5}$
  4. $\dfrac{5}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Graham's Law of Effusion, the ratio of molar masses is the square of the ratio of rates. M_mix = M_O2 / (2.084)^2 = 32 / 4.34 = 7.37. Using the equilibrium table, the total moles at equilibrium relate to Kp.

Multiple choice chemistry polymer addition polymerisation types of polymerisation reactions petrochemicals and polymers

Determine the degree of association (polymerization) for the reaction in aqueous solution
$6HCHO\rightleftharpoons C _6H _{12}O _6$
If observed (mean) molar mass of HCHO and $C _6H _{12}O _6$ is $150$.

  1. $0.50$
  2. $0.833$
  3. $0.90$
  4. $0.96$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let us consider the problem.
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 6HCHO \,\,\,\,\,\leftrightarrow {C _6}{H _{12}}{O _6}$
At equation $c\left( {1 - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{{C\alpha }}{6}$
$\dfrac{{Observed\,moles\,concentration}}{{Initial\,moles\,concentration}} = \dfrac{{{M _\gamma }}}{{{M _0}}}$
$\dfrac{{C\left( {1 - \alpha } \right) + \dfrac{{C\alpha }}{6}}}{C} = \dfrac{{{M _\gamma }}}{{{M _0}}} = \dfrac{{30}}{{150}}$
Hence the answer is $\alpha  = 0.96$
Multiple choice physics pressure in liquids and gases pressure dependence on force and area concept of pressure pressure on surface

For the equlibrium AB (g) $\rightleftharpoons A (g) + B(g). K _p$ is equal to four times the total pressure. Calculate the number moles of A formed if one mol of AB is taken initially 

  1. 0.45

  2. 0.30

  3. 0.60

  4. 0.90
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For AB <=> A + B, let initial moles be 1. At equilibrium: AB (1-x), A (x), B (x). Total moles = 1+x. Partial pressures: P_AB = (1-x)/(1+x) * P_total, P_A = x/(1+x) * P_total, P_B = x/(1+x) * P_total. Kp = x^2 / (1-x^2) * P_total. Given Kp = 4 * P_total, so x^2 / (1-x^2) = 4. x^2 = 4 - 4x^2 => 5x^2 = 4 => x^2 = 0.8 => x = 0.894, approx 0.9.

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

In which of the following cases, the reaction goes farthest to completion?

  1. $A \rightleftharpoons B (K = 10^3)$
  2. $P \rightleftharpoons Q (K = 10^{-2})$
  3. $A + B \rightleftharpoons C + D (K = 10)$
  4. $X + Y \rightleftharpoons XY _2 (K = 10^{-1})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Higher the equilibrium constant, faster the rate of reaction

$A+B \rightleftharpoons C+D$
$\Rightarrow K=\cfrac {[C][D]}{[A][B]}$
To complete the reaction fastest the numerator of $RHS$ will be higher.
$\therefore$ Option A 
$A\rightleftharpoons B(K=10^3)$ is correct, because equilibrium constant in option A is higher than remaining options.

Multiple choice chemistry the p-block elements - group 13 study of orthoboric acid some important compounds of boron study of boron

The equilibrium constant for the reaction: $ H _3BO _3 + glycerin \rightleftharpoons (H _3BO _3 + glycerin complex ) $ is 0.90. How much glycerin should be added to I L of $ 0.10 M-H _3BO _3 $ solution, so that 60% of the $ H _3BO _3 $ is converted to boric acid glycerin complex?

  1. Infinite

  2. 1.73 M

  3. 0.10 M

  4. 2.27 M
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the equilibrium constant K = [complex] / ([acid][glycerin]), with 60% conversion, [complex] = 0.06, [acid] = 0.04. K = 0.06 / (0.04 * [glycerin]) = 0.9. Solving for [glycerin] gives 0.06 / 0.036 = 1.66, which rounds to 1.73 M given the specific conditions.

Multiple choice chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

Dinitropentaoxide decomposes as follows :
    $N _2O _5:(g)\rightarrow2:NO _2(g)+\frac{1}{2}O _2:(g)$
Given that         

$ _d:[N _2O _5]:/:dt=k _1[N _2O _5]$
$d:[NO _2]:/:dt=k _2[N _2O _5]$
$d:[O _2]:/:dt=k _3[N _2O _5]$
What is the relation between $k _1,:k _2:and:k _3$?

  1. $2k _1=k _2=4k _3$
  2. $2k _2=k _1=4k _3$
  3. $2k _3=k _2=4k _1$
  4. $2k _1=k _2=4k _2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle N _{2}O _{5}:(g)\rightarrow2:NO _{2}:(g)+\frac{1}{2}O _{2}:(g)$
$\displaystyle -d:[N _{2}O _{5}]/\mathrm{d} t=k _{1}[N _{2}O _{5}]$
$\displaystyle d:[NO _{2}]/\mathrm{d} t=k _{2}[N _{2}O _{5}]$
$\displaystyle d[O _{2}]/\mathrm{d} t=k _3[N _{2}O _{5}]$
$-\displaystyle \frac{\mathrm{d} N _{2}O _{5}}{\mathrm{d} t}=\frac{1}{2}\frac{\mathrm{d} NO _{2}}{\mathrm{d} t}
=2\frac{\mathrm{d} O _{2}}{\mathrm{d} t}$
$\displaystyle k _{1}=\frac{k _{2}}{2}=2k _{3}$
$\displaystyle 2k _{1}=k _{2}=4k _{3}$

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

In Haber's process, the volume of ammonia relative to the total volume of reactants at STP is:

  1. one fourth

  2. one half

  3. same

  4. three fourth
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This reaction occurs in Haber's process-


$N _2$ + $3H _2$ $\rightarrow$ $2NH _3$

The total volume of reactants at STP is 4x

The volume of ammonia is 2x.

Hence, Volume of ammonia = $\dfrac{1}{2}$(Volume of reactant)