Chemistry

Chemical Reactions and Equations

684 Questions

Chemical reactions and equations form a foundational chemistry topic where students identify reaction products, balance chemical formulas, and classify reaction types. It covers critical mechanisms like Markovnikov's rule, precipitation, and endothermic or exothermic processes. These questions are highly common in general science sections of state and central government competitive exams.

Endothermic reactionsPrecipitation reactionsDehydrohalogenation of alkyl halidesChemical equation balancingReaction product identification

Chemical Reactions and Equations Questions

Multiple choice chemistry rocks and minerals petroleum properties of alkanes petroleum and natural gas

Ethyl bromide reacts with sodium lead alloy to form?

  1. Tetraethyl lead

  2. Ethyl sodium

  3. Ethane

  4. Wthene
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When we treat ethyl bromide with lead sodium alloy we get tetra ethyl lead [TEL]


$4NaPb+4CH _3CH _2Br\longrightarrow \overset {\uparrow Tetraethyllead}{(CH _3CH _2) _4Pb}+4NaCl+3Pb$

TEL is used to increase their efficiency and performance.

Multiple choice chemistry principles of metallurgy concentration of ore concentration of ores general principles of metallurgy

Which of the following reactions does not take place during leaching for concentration of bauxite?

  1. $Al _2O _3+2NaOH+3H _2O\rightarrow 2Na[Al(OH) _4]$
  2. $2Na[Al(OH) _4]+CO _2\rightarrow Al _2O _3.xH _2O+2NaHCO _3$
  3. $Al _2O _3.2H _2O\xrightarrow{{\Delta}}Al _2O _3+2H _2O$
  4. $Al _2O _3.xH _2O\xrightarrow{{\Delta}}Al _2O _3+xH _2O$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
  • Bauxite is the principal ore of aluminium.  The impurities present in bauxite are $SiO _2$, iron oxides and titanium oxide ($TiO _2$). 
  • Concentration is carried out by heating the powdered ore with a concentrated solution of NaOH at 473 – 523 K and 35 – 36 bar pressure. This process is called digestion. 
  •  $Al _2O _3$ is extracted out as sodium aluminate. The impurity, $SiO _2$ too dissolves forming sodium silicate. Other impurities are left behind.
$Al _2O _3+2NaOH+3H _2O \rightarrow 2Na[Al(OH) _4]$
  • The sodium aluminate present in solution is neutralised by passing $CO _2$ gas and hydrated $Al _2O _3$ is precipitated. At this stage, small amount of freshly prepared sample of hydrated$Al _2O _3$ is added to the solution. This is called seeding. It induces the precipitation.
$2Na[Al(OH) _4]+CO _2 \rightarrow Al _2O _3.xH _2O+2NaHCO _3$
  • Sodium silicate remains in the solution and hydrated alumina is filtered, dried and heated to give back pure$Al _2O _3$.
$Al _2O _3.xH _2O \rightarrow Al _2O _3 + xH _2O$

  • Hence, option C is correct answer.

Multiple choice chemistry principles of metallurgy concentration of ore concentration of ores general principles of metallurgy

In the leaching of gold by cyanide process, the following reaction takes place: 
$\displaystyle aAu+bKCN+cH _{2}O+dO _{2}\rightarrow eKAu(CN) _{2}+fKOH$
$\displaystyle uAu+vKCN+wH _{2}O+xO _{2}\rightarrow yKAu(CN) _{4}+zKOH$
Identify the correct statements with respect to the stoichiometric coefficients when these reactions are balanced using simplest natural numbers.

  1. $d, c, x, u$ are in anthmetic progression
  2. $c, y, w, b$ are in arithmetic progression
  3. $f, b, z, v$ are in geometic progressinon
  4. $d, c,a, b$ are geometnic progression
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

$\displaystyle 4Au+8KCN+2H _{2}O+O _{2}\rightarrow 4KAu(CN) _{2}+4KOH$
$\displaystyle 4Au+16KCN+6H _{2}O+3O _{2}\rightarrow 4KAu(CN) _{4}+12KOH$
$a = 4$      $ b = 8$       $c = 2$      $ d = 1$        $e = 4$               $f = 4$
$u = 4$     $ v = 16$       $ w = 6$       $ x = 3$     $ y = 4$           $ z = 12$ 

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

Diborane undergoes cleavage and gives aduct with
(i) CO
(ii) $ H _2O $
(iii) $ N(Me) _3 $
(iv) $ NH _3 $

  1. i, ii

  2. i, iii

  3. ii, iii

  4. i, iv
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Diborane (B2H6) undergoes symmetrical or unsymmetrical cleavage with various Lewis bases. It forms adducts with CO and NH3, among others. Option A is the most standard representation of these reactions.

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

Diborane combines with ammonia at ${120}^{o}C$ to give:

  1. ${B} _{2}{H} _{6}.{NH} _{3}$
  2. ${B} _{2}{H} _{6}.2{NH} _{3}$
  3. ${B} _{2}{H} _{6}.3{NH} _{3}$

  4. none of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Diborane combines with ammonia to form a borazine.

${ B } _{ 2 }{ H } _{ 6 }+2N{ H } _{ 3 }\rightarrow { B } _{ 2 }{ H } _{ 6 }.2N{ H } _{ 3 }$
The reaction proceeds further to form inorganic benzene.
$2{ B } _{ 2 }{ H } _{ 6 }.2N{ H } _{ 3 }\rightarrow 2{ B } _{ 3 }{ N } _{ 3 }{ H } _{ 6 }+12{ H } _{ 2 }$
Here, ${ B } _{ 3 }N _{ 3 }{ H } _{ 6 }$ is inorganic benzene.

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

Borazine is sometimes called inorganic benzene. Which of the following reactions is expected to give this compound?

  1. $B _{2}H _{6}+NH _{3} \xrightarrow [ excess\ { NH } _{ 3 } ]{ low\ tem } $
  2. $B _{2}H _{6}+NH _{3} \xrightarrow [ excess\ { NH } _{ 3 } ]{ high\ tem } $
  3. $B _{2}H _{6}+NH _{3} \xrightarrow [ ratio\ 2\ { NH } _{ 3 }:1{ B } _{ 2 }{ H } _{ 6 } ]{ high\ tem } $
  4. $All\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Borazine is an inorganic compound with the chemical formula ${ (BH } _{ 3 }{ )(NH } _{ 3 })$ In this cyclic compound, the three BH units and three NH units alternate. The compound is isoelectronic and isostructural with benzene. Like benzene, borazine is a colourless liquid. For this reason, borazine is sometimes referred to as "inorganic benzene"

Borazine is synthesized from diborane and ammonia in a 1:2 ratio at 250–300 °C with a conversion of 50%.

$3B _2H _6 + 6NH _3 \rightarrow 2B _3H _6N _3 + 12H _2$

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

$BCl _3+LiAlH _4\rightarrow A+LiCl+AlCl _3$
$A+H _2O\rightarrow B+H _2$
$B\rightarrow C$.
in this reaction sequence A, B, and C compounds respectively are?

  1. $B _2H _6,B _2O _3,B$
  2. $B _2H _6,H _3BO _3,B _2O _3$
  3. $B _2H _6,H _3BO _3,B$
  4. $HBF _4,H _3BO _3,B _2O _3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The reactions given are,

$LiAlH _4 + BCl _3 \rightarrow B _2H _6 + LiCl + AlCl _3$
Hence A is $B _2H _6$

Now, second reaction is,
$B _2H _6 +H _2O \rightarrow H _3BO _3 +H _2$
Hence B is $H _3BO _3$

On heating B i.e. $H _3BO _3$ we get $B _2O _3$

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

$B _2H _6 + NH _3 \rightarrow$ Addition compound $(X)$
$(X) \overset{450K}{\rightarrow} Y + Z(g)$
In the above sequence $Y$ and $Z$ are respectively:

  1. borazine, $H _2O$
  2. boron, $H _2$
  3. boron nitride, $H _2$

  4. borazine and hydrogen
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$B _2H _6+2NH _3 \longrightarrow \underset {(X)}{B _2H _62NH _3}$

$3B _2H _62NH _3\xrightarrow[]{450K}\underset {(Y)}{2B _3N _3H _6}+\underset {(Z)}{3H _2}$
$(Y)$ & $(Z)$ are borazine and $H _2$ respectively.

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

The product obtained when one mole of diborane reacts with two mole of $NH _3 $ at high temperature ?

  1. $B _2H _6.2NH _3 $
  2. $B _3N _3H _6 $
  3. $(BN) _x $
  4. $ [BH _2 (NH _3) _2]^ + BH _4^- $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Diborane react with ammonia
$3B _2H _6 + 6NH _3 → 2B _3N _3H _6 + 12H _2$
So the ratio of combination of $B _2H _6$ and $NH _3$ is $1:2$
Hence option B is correct answer.
Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

On hydrolysis, diborane produces:

  1. ${ H } _{ 3 }BO _{ 2 }+H _{ 2 }{ O } _{ 2 }$
  2. ${ H } _{ 3 }BO _{ 3 }+H _{ 2 }$
  3. $B _{ 2 }O _{ 3 }+{ O } _{ 2 }$
  4. $H _{ 3 }BO _{ 3 }+{ H } _{ 2 }{ O } _{ 2 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Diborane react with water to give following products,
$B _2H _6 + 6H _2O → 2H _3BO _3 + 6H _2$
Hence option $B$ is correct answer.
Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

In the reaction, $2X+B _{2}H _{6}\rightarrow [BH _{2}(X) _{2}]^{+}BH _{4}^{-}$, the amine(s) X is (are) :

  1. $NH _{3}$
  2. $CH _{3}NH _{2}$
  3. $(CH _{3}) _2NH$
  4. $(CH _{3}) _{3}N$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

Smaller amines such as $NH _3$, $CH _3NH _2$ and $(CH _3) _2NH$ give unsymetrical cleavage of diborane according to following reaction
$2NH _3+B _{2}H _{6}\rightarrow [BH _{2}(NH _3) _{2}]^{+}BH _{4}^{-}$
Large amines such as $(CH _3) _3N$ give symmetrical cleavage of diborane according to following reaction
$B _2H _6 + 2N(CH _3) _3 \rightarrow 2(CH _3) _3N \rightarrow BH _3$

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

Diborane is formed from the elements as shown in equation (1) $ 2B(s) +3H _2(g) \rightarrow B _2H _6(g) ......(1) $
The $ H^o$ for the reaction (1) is 
Given that 

$ { H } _{ 2 }O(\ell )\rightarrow H _{ 2 }O(g) $ $ \Delta { H } _{ 1 }^{ 0 }=44kJ $
$ 2B(s)+\frac { 3 }{ 2 } { O } _{ 2 }(g)\rightarrow B _{ 2 }O _{ 3 }(s) $ $ \Delta { H } _{ 2 }^{ 0 }=-1273kJ $
$ B _{ 2 }H _{ 6 }(g)+3O _{ 2 }(g)\rightarrow B _{ 2 }O _{ 3 }+3H _{ 2 }O(g) $ $ \Delta { H } _{ 3 }^{ 0 }=-2035kJ $
$ H _{ 2 }(g)+\frac { 1 }{ 2 } O _{ 2 }(g)\rightarrow H _{ 2 }O(\ell ) $ $ \Delta { H } _{ 4 }^{ 0 }=-286kJ $

  1. 36 kJ

  2. 509 kJ

  3. 520kJ

  4. -3550kJ
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using Hess's Law, we combine the given equations to find the enthalpy of formation of B2H6. The calculation results in -3550 kJ based on the provided thermochemical data.