Chemistry

Chemical Equilibrium and Stoichiometry

135 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 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 the s-block elements (alkali and alkaline earth metals) anomalous properties of lithium group 1 elements: alkali metals properties of s block elements

$LiOH$ reacts with $CO {2}$ to form $Li _{2}CO _{3}$ (atomic mass of $Li = 7)$. The amount of $CO _{2}$ (in g) consumed by $1\ g$ of $LiOH$ is closest to ______.

  1. $0.916$
  2. $1.832$
  3. $0.544$
  4. $1.088$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Balanced chemical equation is:

$2LiOH+C{ O } _{ 2 }\rightarrow { Li } _{ 2 }C{ O } _{ 3 }+{ H } _{ 2 }O$
Molecular weight of each specie is:
$LiOH$=24 g/mol; $C{ O } _{ 2 }$=44 g/mol; ${ Li } _{ 2 }C{ O } _{ 3 }$= 74 g/mol; ${ H } _{ 2 }O$=18 g/mol
From the reaction:
2 mol of $LiOH$ ($24\times2=48\ g$) consumes 1 mol of $CO _2$ (44 g)
Therefore, 1 g of $LiOH$ will consume $\frac { 1 }{ 48 } \times 44\quad g\quad C{ O } _{ 2 }=0.916\ g\ CO _2$
Option A is the correct answer.

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

The reaction, $Sucrose\xrightarrow [  ]{ { H }^{ + } } Glucose+Fructose$, takes  place at certain temperature while the volume of solution is maintained at $1$ litre. At time zero the initial rotation of the mixture is ${ 34 }^{ o }C$.After $30$ minutes the total rotation of solution is ${ 19 }^{ o }C$ and after a very long time, the total rotation is ${ -11 }^{ o }C$. Find the time when solution was optically inactive?

  1. $135$ min
  2. $103.7$ min
  3. $38.7$ min
  4. $45$ min
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

rate constant $k = \dfrac{2.303}{t}log\dfrac{(r _0 - r _\infty) }{(r _t-r _\infty)} = 0.0135$
At the point of optical inactiveness, rotation is zero. 

So, time taken is $ t =\dfrac{ 2.303}{k}log(45/11) = 103.7$ min.

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 chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

Law of multiple proportions can be illustrated by taking the example of:

  1. $NaOH$ and $KOH$
  2. $NaCl$ and $NaBr$
  3. $SO _2$ and $SO _3$
  4. $H _2CO _3$ and $CO _2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Law of multiple proportions, statement that when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.

$SO _2$ and $SO _3$ fulfiling this as the weight of oxygen is in ratio of $\dfrac{2}{3}$.

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

$P _xO _y$ what will be correct value of $x$ and $y$ if $P$ and $H$ combine in the mass ratio of $3.1 : 0.3$ and in water $H$ and $O$ combine in the mass ratio $0.2 : 1.6$?

  1. $1$ and $1$
  2. $1$ and $2$
  3. $2$ and $3$
  4. $2$ and $5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 

$P:H = 3.1 : 0.3 = (3.1 \times : 0.30) \times 2 = 6.2 : 0.6$

$H:O = 0.2 : 1.6 = (0.2 \times 3 : 1.6) \times 3 = 0.6 : 4.8$

$P:O = 6.2 : 4.8 = 3.1 : 2.4 $ (ratio by mass) $= 1.29: 1$

$1.29$ mass $P= 1 $ mass $O$

$x$ mass $P= 16 \ g$ mass $O$

$x= 16 \times 1.29 = 20.64 \ g$ Phosphorous

$= \cfrac {20.64}{31}$ moles Phosphorous

$= 0.66$

$1 \ mole \ O(y) = 0.66 \ mole \ P(x) = \cfrac 23$

$\therefore \ x:y = 2:3 $

$x =2 \ ; \ y=3$

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)

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

In the reaction, $N _2 + 3H _2 \rightarrow 2NH _3$, the ratio of volumes of nitrogen, hydrogen and ammonia is 1 : 3: 2. These figures illustrate the law of: 

  1. constant proportions

  2. Gay-Lussac

  3. multiple proportions

  4. reciprocal proportions
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

ANS; B
Gay Lussac’s Law of Combining Volumes states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the product(s) formed if gaseous, provided the temperature and pressure remain constant.

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

Which of the following statements is/are correct?

  1. A sample of $CaCO _3$ contains $Ca = 40\%,\ C = 12\%$ and $O= 48\%$. If the law of constant composition is true, then the mass of $Ca$ in $10$ g of $CaCO _3$ from another source is $4.0$ g.
  2. $12$ g of carbon is heated in vacuum and there is no change in the mass. This is the best example of the law of conservation of mass.

  3. Air is heated at constant pressure and there is no change in mass but the volume increases. This is the best example of the law of conservation of mass.
  4. $SO _2$ gas was prepared by (i) heating $Cu $ with conc. $H _2SO _4$ (ii) burning sulphur in oxygen (iii) reacting sodium sulphite $(Na _2SO _3)$ with dilute $H _2SO _4$. It was observed that is each case, $S$ and $O$ combine in the ratio of $1:1$. This data illustrates the law of constant composition.
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

(A) A sample of $CaCO _3$ contains $Ca = 40\%$, $C = 12\%$ and $O= 48\%$. If the law of constant composition is true, then the mass of $Ca$ in $10$ g of $CaCO _3$ from another source is $\dfrac {40 \times 10 }{100}= 4.0 $ g


According to the law of constant composition, all samples of a given chemical compound have the same elemental composition by mass.
Hence, the statement A is true.

(B) $12$ g of carbon is heated in vacuum and there is no change in the mass. This is not the best example of the law of conservation of mass as there is no chemical transformation involved since there is a vacuum.
The Law of Conservation of Mass states that matter can be changed from one form into another, mixtures can be separated or made and pure substances can be decomposed but the total amount of mass remains constant.
Hence, the statement B is false.

(C) Air is heated at constant pressure and there is no change in mass but the volume increases. This is not the best example of the law of conservation of mass as there is no chemical transformation involved since there is a vacuum.

The Law of Conservation of Mass states that matter can be changed from one form into another, mixtures can be separated or made and pure substances can be decomposed but the total amount of mass remains constant.
Hence, the statement C is false.

(D) $SO _2$ gas was prepared by (i) heating $Cu$ with conc. $H _2SO _4$ (ii) burning sulphur in oxygen (iii) reacting sodium sulphite $(Na _2SO _3)$ with dilute $H _2SO _4$. It was observed that is each case, S and O combine in the ratio of $1:1$. This data illustrates the law of constant composition.

According to the law of constant composition, all samples of a given chemical compound have the same elemental composition by mass.

Hence, the statement D is true.