$A{\rm{ }}is{\rm{ }}C{H_3}COOH{\rm{ }}and{\rm{ }}B{\rm{ }}is{\rm{ }}C{H_3}COONa$

On decarboxylation with soda lime it gives Methane

On Kolbe’s electrolysis it gives Ethane

On decarboxylation with soda lime it gives Ethane

The correct Option is "Both 1 and 2"

The electrons in the reaction centre of PS-I are

Excited simultaneously with PS-II (${P_{680}}$). The movement of electrons in ETS of photosynthesis is down hill in terms of oxidation reduction or redox potential scale. The electrons are not used up as they pass through the electron transport chain, but they passed on the pigments of photosystem I. Simultaneously, electrons in the reaction centre of PS-I are also excited, when they received red light of wavelength 700 nm and are transferred to another acceptor molecule that has greater redox potential. These electrons than are moved down hill again this time to a molecule of energy rich NADP^+. The addition of these electrons reduces the $NAD{P^ + }$ to $NADPH + {H^ + }$

Excited simultaneous with PS-II

Excited simultaneously with ${P_{680}}$

Excited simultaneously with ${P_{700}}$

The correct Option is "Either (a) or (b)"

The molecule present in the reaction centre of photosystem is

Chlorophyll-a. The pigments are organised into two discrete photochemical Light Harvesting Completes (LHC) within the Photosystem I (PS-I) and Photosystem II (PS-I). These are named in the sequence of their discovery, and not in the sequence in which they function during the light reaction. The LHC are made up of hundreds of pigment molecules bounded to proteins. Each photosystem has all the pigments (except one molecule of chlorophyll-a) forming a light harvesting system known as antenna

The correct Option is "Chlorophyll-a"

In the reaction of ${{\rm{O}}_3}$ and ${{\rm{H}}_2}{{\rm{O}}_2},$ the later acts as :

${{\rm{O}}_3} + {{\rm{H}}_2}{{\rm{O}}_2} \to {{\rm{H}}_2}{\rm{O}} + 2{{\rm{O}}_2};{{\rm{H}}_2}{{\rm{O}}_2}$ is reduced

The correct Option is "Oxidising agent"

Both oxidising and bleaching agent

The eq. wt. of ${\rm{N}}{{\rm{a}}_2}{{\rm{S}}_2}{{\rm{O}}_3}$ as reductant, in the reaction, ${\rm{N}}{{\rm{a}}_2}{{\rm{S}}_2}{{\rm{O}}_3} + 5{{\rm{H}}_2}{\rm{O}} + 4{\rm{C}}{{\rm{l}}_2} \to 2{\rm{NaHS}}{{\rm{O}}_4} + 8{\rm{HCl}}:$

${\rm{S}}_2^{2 + } \to 2{{\rm{S}}^{6 + }} + 8e$

$\left( {{\rm{Mol}}{\rm{.wt}}{\rm{.}}} \right)/1$

The correct Option is "(Mol. wt.)/8"

In the reaction ; $2{\rm{Ag}} + 2{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} \to {\rm{A}}{{\rm{g}}_2}{\rm{S}}{{\rm{O}}_4} + 2{{\rm{H}}_2}{\rm{O}} + {\rm{S}}{{\rm{O}}_2},{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$act as :

$2e + {S^{6 + }} \to {S^{4 + }}\,S\,of\,{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}\,is\,\,reduced$

The correct Option is "Oxidising agent"

In the reaction, ${{\rm{I}}_2} + 2{{\rm{S}}_2}{\rm{O}}_3^{2 - } \to 2{{\rm{I}}^ - } + {{\rm{S}}_4}{\rm{O}}_6^{2 - },$ Equivalent weight of iodine will be equal to:

${\rm{I}}_2^0 \to 2{{\rm{I}}^ - } + 2{\rm{e}}$

The correct Option is "1/2 of its molecular weight"

In the reaction, ${{\rm{N}}_2} \to {\rm{N}}{{\rm{H}}_3}{\rm{.}}$. The eq.wt. of ${{\rm{N}}_2}$ and ${\rm{N}}{{\rm{H}}_3}$ are respectively equal to :

$6e + {\left( {{N^0}} \right)_2} \to 2{N^{ - 3}}$,$\therefore {E_{{N_2}}} = \frac{{28}}{6};{E_{N{H_3}}} = \frac{{17}}{3}$

$\frac{{28}}{3},\frac{{17}}{3}$

The correct Option is "$\frac{{28}}{6},\frac{{17}}{3}$"

$\frac{{28}}{2},\frac{{17}}{2}$

$\frac{{28}}{5},\frac{{17}}{5}$

The equivalent weight of ${\rm{SnC}}{{\rm{l}}_2}$ in the reaction, ${\rm{SnC}}{{\rm{l}}_2} + {\rm{C}}{{\rm{l}}_2} \to {\rm{SnC}}{{\rm{l}}_4}$ is :

${\rm{S}}{{\rm{n}}^{2 + }} \to {\rm{S}}{{\rm{n}}^{4 + }} + 2e$,$\therefore E = M/2 = \frac{{119 + 71}}{2} = 95$

Which is the best description of behaviour of bromine in the reaction given below? ${{\rm{H}}_2}{\rm{O}} + {\rm{B}}{{\rm{r}}_2} \to {\rm{HBr}} + {\rm{HOBr}}$

In reaction 0 +1 -1 ${{\rm{H}}_2}{\rm{O}} + {\rm{B}}{{\rm{r}}_2} \to {\rm{HOBr}} + {\rm{HBr}}$ The oxidation number of bromine increases from 0 to $ + 1$ and decreases from 0 to $ - 1$, so due to this reason bromine is both oxidised as well as reduced in the above reaction.

The correct Option is "Both oxidised and reduced"

In the reactions;${\rm{A}}{{\rm{s}}_2}{{\rm{S}}_3} + {\rm{HN}}{{\rm{O}}_3} \to {{\rm{H}}_3}{\rm{As}}{{\rm{O}}_4} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} + {\rm{NO}}$, the element oxidized is/ are :

${\rm{A}}{{\rm{s}}^{3 + }} \to {\rm{A}}{{\rm{s}}^{5 + }} + 2e$,${{\rm{S}}^{2 - }} \to {{\rm{S}}^{6 + }} + 8e$

The correct Option is "As and S both"

The eq. wt. of ${\rm{KMn}}{{\rm{O}}_4}$ in the reaction, ${\rm{MnO}}_4^ - + {\rm{M}}{{\rm{n}}^{2 + }} + {{\rm{H}}_2}{\rm{O}} \to {\rm{Mn}}{{\rm{O}}_2} + {{\rm{H}}^ + }$ (unbalanced) is :

$3e + {\rm{M}}{{\rm{n}}^{7 + }} \to {\rm{M}}{{\rm{n}}^{4 + }};E = \frac{{158}}{3} = 52.66$

In the reaction, ${\rm{S}}{{\rm{O}}_2} + 2{{\rm{H}}_2}{\rm{S}} \to 3{\rm{S}} + 2{{\rm{H}}_2}{\rm{O}}$ the substance that oxidizes is,

${{\rm{H}}_2}\mathop S\limits^{ - 2} \to \mathop S\limits^0 $ The oxidation number of S increases from $ - 2$ to 0 in elemental sulphur and hence, ${{\rm{H}}_2}{\rm{S}}$ gets oxidized

The correct Option is "${{\rm{H}}_2}{\rm{S}}$"

In the reaction, ${\rm{VO}} + {\rm{F}}{{\rm{e}}_2}{{\rm{O}}_3} \to {\rm{FeO}} + {{\rm{V}}_2}{{\rm{O}}_5}{\rm{.}}$. The eq.wt. of ${{\rm{V}}_2}{{\rm{O}}_5}$ is equal to its :

$2{V^{2 + }} \to V_2^{5 + } + 6e$

The correct Option is "$mol.{\rm{ }}wt./6$"

In the reaction, ${\rm{C}}{{\rm{r}}_2}{\rm{O}}_7^{2 - } + 14{{\rm{H}}^ + } + 6{{\rm{I}}^ - } \to 2{\rm{C}}{{\rm{r}}^{3 + }} + 3{{\rm{H}}_2}{\rm{O}} + 3{{\rm{I}}_2}$, The eq.wt. of ${\rm{C}}{{\rm{r}}^{3 + }}$) is :

$6e + Cr_2^{6 + } \to 2C{r^{3 + }};{\rm{Eq}}{\rm{.wt}}{\rm{. of Cr}} = \frac{{{\rm{at}}{\rm{.wt}}{\rm{.}}}}{3}$

$\frac{{{\rm{mol}}{\rm{.wt}}{\rm{.}}}}{3}$

$\frac{{{\rm{at}}{\rm{.wt}}{\rm{.}}}}{6}$

The correct Option is "$\frac{{{\rm{at}}{\rm{.wt}}{\rm{.}}}}{3}$"

$\frac{{{\rm{mol}}{\rm{.wt}}{\rm{.}}}}{6}$

In the reaction, ${{\rm{H}}_2}{{\rm{O}}_2} + {\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3} \to {\rm{N}}{{\rm{a}}_2}{{\rm{O}}_2} + {\rm{C}}{{\rm{O}}_2} + {{\rm{H}}_2}{\rm{O}}$ the substance undergoing oxidation is

None of the elements changes its oxidation number.$\mathop {{H_2}{O_2}}\limits^{ - 1} + \mathop {N{a_2}C{O_3}}\limits^{ + 2\,\,\, + 4 - 2} \to \mathop {N{a_2}{O_2}}\limits^{ + 1\,\,\, - 1} + \mathop {C{O_2}}\limits^{ + 4\,\, - 2} + {H_2}\mathop O\limits^{ - 2} $

${\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}$

${\rm{N}}{{\rm{a}}_2}{{\rm{O}}_2}$

The correct Option is "None of these"

In the reaction, ${\rm{Zn}} + 2{{\rm{H}}^ + } + 2{\rm{C}}{{\rm{l}}^ - } \to {\rm{Z}}{{\rm{n}}^{2 + }}2{\rm{C}}{{\rm{l}}^ - } + {{\rm{H}}_2},$ the spectator ion is :

The species present in solution but does not take part in the reaction and are also omitted while writing the potential redox change are called spectator ion.

The correct Option is "${\rm{C}}{{\rm{l}}^ - }$"

What weight of ${\rm{HN}}{{\rm{O}}_3}$ is required to make 1 litre of 2 N solution to be used as an oxidising agent in the reaction? $3{\rm{Cu}} + 8{\rm{HN}}{{\rm{O}}_3} \to 3{\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_3}} \right)_2} + 2{\rm{NO}} + 4{{\rm{H}}_2}{\rm{O}}$

${\rm{Meq}}{\rm{.of HN}}{{\rm{O}}_3} = 1000 \times 2 = 2000$,$\therefore \frac{w}{{63/3}} \times 1000 = 2000$,$\therefore w = 42{\rm{g}}$

The heat evolved during neutralisation is maximum in the reaction of :

Strong acid (HCl) and strong base (NaOH).

${\rm{N}}{{\rm{H}}_4}{\rm{OH}}\,\,\,{\rm{and}}\,\,\,{\rm{C}}{{\rm{H}}_3}{\rm{COOH}}$

${\rm{N}}{{\rm{H}}_4}{\rm{OH}}\,\,\,{\rm{and}}\,\,\,{\rm{HCl}}$

${\rm{NaOH}}\,\,{\rm{and}}\,\,{\rm{C}}{{\rm{H}}_3}{\rm{COOH}}$

The correct Option is "${\rm{NaOH}}\,\,{\rm{and}}\,\,{\rm{HCl}}$"

In the reaction, ${{\rm{H}}_2}\left( {\rm{g}} \right) + {\rm{C}}{{\rm{l}}_2}\left( {\rm{g}} \right)\leftrightharpoons2{\rm{HCl}}\left( {\rm{g}} \right)$

${{\rm{H}}_2}\left( {\rm{g}} \right) + {\rm{C}}{{\rm{l}}_2}\left( {\rm{g}} \right)\leftrightharpoons2{\rm{HCl}}\left( {\rm{g}} \right)$ We know that, ${K_p} = {K_c}.{\left( {RT} \right)^{\Delta {n_g}}}$ $\Delta {n_g}$ = no. of moles of gaseous products – no. of moles of gaseous reactants $\; = 2 - 2 = 0$ ${K_p} = {K_c}.{\left( {RT} \right)^0}$ ${K_p} = {K_c}$

The correct Option is "${K_p} = {K_c}$"

In the reactions, ${\rm{PC}}{{\rm{l}}_5}\leftrightharpoons{\rm{PC}}{{\rm{l}}_3} + {\rm{C}}{{\rm{l}}_2},$, the amounts of ${\rm{PC}}{{\rm{l}}_5},{\rm{PC}}{{\rm{l}}_3}$ and ${\rm{C}}{{\rm{l}}_2}$ at equilibrium are $2$ mole each and the total pressure is $3$ $atm$. The equilibrium constant ${K_p}$ is :

${K_p} = \frac{{^n{\rm{C}}{{\rm{l}}_2}{ \times ^n}{\rm{PC}}{{\rm{l}}_3}}}{{^n{\rm{PC}}{{\rm{l}}_5}}} \times {\left[ {\frac{p}{{\sum n}}} \right]^1}$ $ = 2 \times \frac{2}{2} \times {\left[ {\frac{3}{6}} \right]^1}$ $ = {\rm{ }}1\;{\rm{ }}atm$

The correct Option is "$1.0{\rm{ }}\;atm$"

The number of electrons involved in the reaction when a faraday of electricity is passed through an electrolyte in solution is :

The correct Option is "$6.02 \times {10^{23}}$"

In the reaction $_1^2H + \,_1^3\,\,H \to _2^4He + \,_0^1\,\,n$ if the binding energies of $_1^2H,_1^3\,H$ and $_2^4He$ are respectively $a,b$ and $c$ (in $MeV$), then the energy (in $MeV$) released in this reaction is

During fusion binding energy of daughter nucleus is always greater than the total energy of the parent nuclei so energy released $ = c - \left( {a + b} \right) = c - a - b$

The correct Option is "$c - a - b$"

In the reaction $_7{{\rm{N}}^{14}} + \alpha { \to _8}{X^{17}}{ + _1}{p^1}$ identify $X$.

In the reaction $\;\;\;\;\;{\;_7}{N^{14}} + \,\,\alpha { \to _8}{X^{17}}{ + _1}{p^1}$ ${\rm{8}}$ is the atomic number of oxygen molecule. So, here $X$ is oxygen$\left( {{{\rm{O}}_2}} \right)$ molecule.

The correct Option is "${{\rm{O}}_2}$"

${{\rm{N}}_2}\left( {\rm{g}} \right) + 3{{\rm{H}}_2}\left( {\rm{g}} \right)\;\leftrightharpoons\;2{\rm{N}}{{\rm{H}}_3}\left( {\rm{g}} \right)$ In the reaction given above, the addition of small amount of an inert gs at constant pressure will shift the equilibrium towards which side?

In the given equilibrium reaction, if inert gas is added at constant pressure, it will result in increased volume. Due to which, the equilibrium will shift towards the left hand side (LHS) i.e., reaction goes in back direction.

The correct Option is "LHS (Left hand side)"

In the reaction, ${\rm{PC}}{{\rm{l}}_5}\left( {\rm{g}} \right)\leftrightharpoons{\rm{PC}}{{\rm{l}}_3}\left( {\rm{g}} \right) + {\rm{C}}{{\rm{l}}_2}\left( {\rm{g}} \right),$ the equilibrium concentration of ${\rm{PC}}{{\rm{l}}_5}\;and\;\;{\rm{PC}}{{\rm{l}}_3}$ are $0.4{\rm{ }}\;and\;{\rm{ }}0.2\;{\rm{ }}mol/L$ respectively. If the value of ${K_c}$ is $0.5$ what is the concentration of ${\rm{C}}{{\rm{l}}_2}$ in $mol/L?$

${K_c} = \frac{{\left[ {{\rm{PC}}{{\rm{l}}_3}} \right]\left[ {{\rm{C}}{{\rm{l}}_2}} \right]}}{{\left[ {{\rm{PC}}{{\rm{l}}_5}} \right]}} = \frac{{0.2 \times x}}{{0.4}} = 0.5$ $\therefore \;x = 1$

The volume of the reaction vessel containing an equilibrium mixture in the reaction, \[S{O_2}C{l_2}\left( g \right)\,\,\leftrightharpoons\,\,S{O_2}\left( g \right){\rm{ }} + {\rm{ }}C{l_2}\left( g \right)\] is increased. When equilibrium is reestablished:

An increase in volume for the equilibrium having, ${K_c} = \frac{{\left[ {{\rm{S}}{{\rm{O}}_2}} \right]\left[ {{\rm{C}}{{\rm{l}}_2}} \right]}}{{\left[ {{\rm{S}}{{\rm{O}}_2}{\rm{C}}{{\rm{l}}_2}} \right]}} = \frac{{\left[ {{\rm{mole of S}}{{\rm{O}}_2}} \right]\left[ {{\rm{mole of C}}{{\rm{l}}_2}} \right]}}{{\left[ {{\rm{mole of S}}{{\rm{O}}_2}{\rm{C}}{{\rm{l}}_2}} \right] \times {\rm{volume}}}}$ Will increase the mole of ${\rm{C}}{{\rm{l}}_2}$ or ${\rm{S}}{{\rm{O}}_2}$ to maintain ${K_c}$ values constant.

The amount of $S{O_2}\left( g \right)$ will decrease

The amount of $S{O_2}C{l_2}\left( g \right)$ will increase

The correct Option is "The amount of $C{l_2}\left( g \right)$ will increase"

The amount of $C{l_2}\left( g \right)$ will remain unchanged

In the reaction,${\rm{C}}\left( s \right) + {\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right)\leftrightharpoons2{\rm{CO}}\left( {\rm{g}} \right),$ the equilibrium pressure is $12{\rm{ }}\;atm.$ If $50\% $ of ${\rm{C}}{{\rm{O}}_2}$ reacts, ${K_p}$ for the change is :

$\begin{array}{l} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{C}}\left( s \right) + {\rm{C}}{{\rm{O}}_2}\leftrightharpoons2{\rm{CO}}\left( {\rm{g}} \right)\;\\ \begin{array}{*{20}{l}} {Gaseous{\rm{ }}mole{\rm{ }}before\;\;\; - \;\;\;\;\;\;\;\;\;\;\;1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0}\\ {dissociation} \end{array}\\ Gaseous{\rm{ }}mole{\rm{ }}after\;\;\;\;\; - \;\;\;\;\;\;\left. {\left( {1 - \frac{{50}}{{100}}} \right.} \right)\left. {\left( {\frac{{2 \times 50}}{{100}}} \right.} \right) \end{array}$ $\begin{array}{l} dissociation\\ 1.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0.5\;\;\;\;\;\;\;\;\;\;\;1 \end{array}$ ∴ Total mole = 1.5 and $\Delta n = 1$ Total pressure given at equilibrium = 12 atm ${K_p} = \frac{{{{\left( {{n_{{\rm{CO}}}}} \right)}^2}}}{{\left( {{n_{{\rm{C}}{{\rm{O}}_2}}}} \right)}} \times {\left[ {\begin{array}{*{20}{c}} {\frac{P}{{\sum n}}} \end{array}} \right]^{\Delta n}} = \frac{{{{\left( 1 \right)}^2}}}{{0.5}} \times {\left. {\left( {\frac{{12}}{{1.5}}} \right.} \right)^1}$ ${K_p} = \frac{{12}}{{1.5 \times 0.5}} = 16\;{\rm{atm}}$

The correct Option is "$16\;atm.$"

In the reaction, $3A + 2B \to 2C$, the equilibrium constant ${K_c}$ is given by

$3A + 2B \to 2C$ ${K_c} = \frac{{{\rm{concentration of products}}}}{{{\rm{concnetration of reactants}}}}$ $ = \frac{{{{\left[ C \right]}^2}}}{{{{\left[ A \right]}^3} \times {{\left[ B \right]}^2}}}$

$\frac{{\left[ {3A} \right] \times \left[ {2B} \right]}}{{\left[ C \right]}}$

$\frac{{{{\left[ A \right]}^3} \times \left[ B \right]}}{{\left[ C \right]}}$

The correct Option is "$\frac{{{{\left[ C \right]}^2}}}{{{{\left[ A \right]}^3} \times {{\left[ B \right]}^2}}}$"

$\frac{{\left[ C \right]}}{{\left[ {3A} \right]\left[ {2B} \right]}}$

In the reactions, $A + 2B\leftrightharpoons2C$, if 2 moles of $A$, 3.0 moles of $B$ and 2.0 moles of C are placed in a 2 L flask and the equilibrium concentration of $C$ is 0.5 mol/L, the equilibrium constant $\left( {{K_c}} \right)$ for the reactions is

$\mathop A\limits_2 + \mathop {2B}\limits_3 \leftrightharpoons\mathop {2C}\limits_2 \;\;initial{\rm{ }}\;moles$ $\left( {2 - 0.5} \right)\left( {3 - 0.5} \right)\left( {2 \times 0.5} \right)$ at equilibrium Molar concentration of $A = \frac{{1.5}}{2}$ Molar concentration of $B = \frac{{2.5}}{2}$ Molar concentration of $C = \frac{1}{2}$ $K = \frac{{{{\left[ C \right]}^2}}}{{\left[ A \right]{{\left[ B \right]}^2}}} = \frac{{1 \times 1 \times 2 \times 2 \times 2 \times 2}}{{2 \times 2 \times 1.5 \times 2.5 \times 2.5}}$ $ = \frac{2}{{1.5 \times 2.5 \times 2.5}} = 0.21$

The number of ${\rm{\alpha }}$-particles and ${\rm{\beta }} - {\rm{particles}}$ respectively emitted in the reaction $_{88}{A^{196}}{ \to _{78}}{B^{164}}$ are

$_{88}{A^{196}}{ \to _{78}}{B^{164}}$ Number of ${\rm{\alpha }} - {\rm{particles}}$, $ = \frac{{196 - 164}}{4} = 8$ $_{88}{A^{196}}\xrightarrow{{ - 8\alpha }}{}_{78}{X^{164}}{ \to _{78}}{B^{164}}$ $\therefore $ Number of ${\rm{\beta }} - {\rm{particles}}$ =$78 - 72 = 6$

$\begin{array}{*{20}{l}} {{\rm{8}}\,\,{\rm{and}}\,\,{\rm{8}}} \end{array}$

The correct Option is "$\begin{array}{*{20}{l}} {{\rm{8}}\,\,{\rm{and}}\,\,{\rm{6}}} \end{array}$"

$\begin{array}{*{20}{l}} {{\rm{6}}\,\,{\rm{and}}\,\,{\rm{8}}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{6}}\,\,{\rm{and}}\,\,{\rm{6}}} \end{array}$

In the reaction ${{\rm{I}}_2} + {{\rm{I}}^ - } = {\rm{I}}_3^ - ,$ the Lewis base is:

${{\rm{I}}^ - }$ is electron pair donor.

The correct Option is "${{\rm{I}}^ - }$"

If in the reaction ${{\rm{N}}_2}{{\rm{O}}_4}\;\;\leftrightharpoons\;\;2{\rm{N}}{{\rm{O}}_2}$, $x$ is that part of ${{\rm{N}}_2}{{\rm{O}}_4}$ which dissociates, then the number of molecules at equilibrium will be

$\begin{array}{l} {{\rm{N}}_2}{{\rm{O}}_4}\leftrightharpoons2{\rm{N}}{{\rm{O}}_2}\\ 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0\;\;\;\;\;\;\;\;\;\;\;\;Before{\rm{ }}\;dissociation\;\;\\ 1 - x\;\;\;\;\;\;\;\;\;\;2x\;\;\;\;\;\;\;\;\;After{\rm{ }}\;dissociation \end{array}$ Total number of moles at equilibrium $ = 1 - x\;\; + 2x\;\; = 1 + x$

The correct Option is "$\left( {1 + x} \right)$"

If in the reaction :${{\rm{N}}_2}{{\rm{O}}_4}\;\leftrightharpoons2{\rm{N}}{{\rm{O}}_2},\;\alpha $ is degree of dissociation of ${{\rm{N}}_2}{{\rm{O}}_4}$, then the number of molecules at equilibrium will be:

$\begin{array}{l} {{\rm{N}}_2}{{\rm{O}}_4}\;\leftrightharpoons2{\rm{N}}{{\rm{O}}_2}\\ \;\;\;\;1\;\;\;\;\;\;\;\;\;\;\;0\\ \;1 - \alpha \;\;\;\;\;\;\;\;2\alpha \end{array}$ $\therefore $ Total mole at equilibrium $ = 1 - \alpha + 2\alpha = 1 + \alpha $

${\left( {1 - \alpha } \right)^2}$

The correct Option is "$\left( {1 + \alpha } \right)$"

In the reaction, ${{\rm{N}}_2}{{\rm{O}}_4}\;\leftrightharpoons\;2{\rm{N}}{{\rm{O}}_2}$, ${\rm{\alpha }}$ is that part of ${{\rm{N}}_2}{{\rm{O}}_4}$ which dissociates, then the number of moles at equilibrium will be

$\begin{array}{l} {{\rm{N}}_2}{{\rm{O}}_4}\;\leftrightharpoons\;2{\rm{N}}{{\rm{O}}_2}\\ \;\;\;\;1\;\;\;\;\;\;\;\;\;\;\;\;\;0\;\;\;\;\;initially\\ \left( {1 - {\rm{\alpha }}} \right)\;\;\;\;\;\;\;\;\;2{\rm{\alpha }}\;\;\;at\;\;equilibrium \end{array}$ Total moles at equilibrium $ = \left( {1 - {\rm{\alpha }}} \right) + 2{\rm{\alpha }}$ $ = 1 + {\rm{\alpha }}$

The correct Option is "$\left( {1 + {\rm{\alpha }}} \right)$"

${\left( {1 - {\rm{\alpha }}} \right)^2}$

Partial pressure of ${{\rm{O}}_2}$ in the reaction $2{\rm{A}}{{\rm{g}}_2}{\rm{O}}\left( s \right)\;\leftrightharpoons\;4{\rm{Ag}}\left( s \right) + {{\rm{O}}_2}\left( {\rm{g}} \right)$

$2{\rm{A}}{{\rm{g}}_2}{\rm{O}}\left( s \right)\;\leftrightharpoons\;4{\rm{Ag}}\left( s \right) + {{\rm{O}}_2}\left( {\rm{g}} \right)$ ${K_p} = {p_{{{\rm{O}}_2}}}$ $({ \because }\;\;\;{\rm{Ag}}\;{\rm{ and }}\;{\rm{A}}{{\rm{g}}_2}{\rm{O }}\;{\rm{are }}\;{\rm{solids}})$

The correct Option is "${K_p}$"

In the reaction $C{H_2} = C{H_2}\xrightarrow{{{\rm{hypochlorous}}\,\,{\rm{acid}}}}?$

$C{H_2} = C{H_2} + HOCl \to \mathop {C{H_2}}\limits_{\mathop |\limits_{Cl} } - \mathop {C{H_2}}\limits_{\mathop |\limits_{OH} } $

The correct Option is "$C{H_2}Cl.C{H_2}OH$"

In the reaction ${\rm{BrO}}_3^ - \left( {aq} \right) + 5{\rm{B}}{{\rm{r}}^ - }\left( {aq} \right) + 6{{\rm{H}}^ + } \to 3{\rm{B}}{{\rm{r}}_2}\left( l \right) + 3{{\rm{H}}_2}{\rm{O}}\left( l \right)$ The rate of appearance of bromine $({\rm{B}}{{\rm{r}}_2})$ is related to rate of disappearance of bromide ions as following:

$\frac{{d\left( {{\rm{B}}{{\rm{r}}_2}} \right)}}{{dt}} = - \frac{3}{5}\frac{{d\left( {{\rm{B}}{{\rm{r}}^ - }} \right)}}{{dt}}$

$\frac{{d\left( {{\rm{B}}{{\rm{r}}_2}} \right)}}{{dt}} = \frac{3}{5}\frac{{d\left( {{\rm{B}}{{\rm{r}}^ - }} \right)}}{{dt}}$

The correct Option is "$\frac{{d\left( {{\rm{B}}{{\rm{r}}_2}} \right)}}{{dt}} = - \frac{3}{5}\frac{{d\left( {{\rm{B}}{{\rm{r}}^ - }} \right)}}{{dt}}$"

$\frac{{d\left( {{\rm{B}}{{\rm{r}}_2}} \right)}}{{dt}} = - \frac{5}{3}\frac{{d\left( {{\rm{B}}{{\rm{r}}^ - }} \right)}}{{dt}}$

$\frac{{d\left( {{\rm{B}}{{\rm{r}}_2}} \right)}}{{dt}} = \frac{5}{3}\frac{{d\left( {{\rm{B}}{{\rm{r}}^ - }} \right)}}{{dt}}$

In the reaction, $2{N_2}{O_5} \to 4N{O_2} + {O_2}$ initial pressure is $500\;atm$ and rate constant $k$ is $3.38 \times {10^{ - 5}}\;{s^{ - 1}}$ after 10 min the final pressure of ${N_2}{O_5}$ is

From the unit of rate constant $(i.e.,\;{s^{ - 1}}),$ it is clear that the reaction is of first order. $2{N_2}{O_5} \to 4N{O_2} + {O_2}$ Hence, for first order reaction , $k = \frac{{2.303}}{t}\log \frac{{{p_0}}}{{{p_t}}}$ $\therefore 3.38 \times {10^{ - 5}} = \frac{{2.303}}{{10 \times 60}}\log \frac{{500}}{{{p_t}}}$ $or\;\log \frac{{500}}{{{p_t}}} = 0.00880$ $\therefore \frac{{500}}{{{p_t}}} = \;anti\;\log \;0.00880$ $ = 1.02$ ${p_t} = \frac{{500}}{{1.02}} = 490\;atm$

The correct Option is "490 atm"

In the reaction, $X\,{\rm{and}}\,Y$ are $H - C \equiv CH\xrightarrow[{ii)C{H_3}C{H_2}Br}]{{i)NaN{H_2}/liq.N{H_3}}}X\xrightarrow[{ii)C{H_3}C{H_2}Br}]{{i)NaN{H_2}/liq.N{H_3}}}Y$

The correct Option is "1 – butyne and 3 - hexyne"

In the reaction sequence, the product $Y$ is ${C_6}{H_6}\xrightarrow[{AlC{l_3}}]{{C{H_3}Cl}}X\xrightarrow{{KMn{O_4}}}Y$

${C_6}{H_6}\xrightarrow[{AlC{l_3}}]{{C{H_3}Cl}}{C_6}{H_5} - C{H_3}\xrightarrow{{KMn{O_4}}}{C_6}{H_5} - COOH$

The correct Option is "benzoic acid"

In the reaction $A + B \to products$, if B is taken in excess, then it is an example of

Pseudo first order reaction Since B is present in large excess.The concentration of B remains practically constant and hence the rate will vary directly with the concentration of A. So it is pseudo first order reaction.

The correct Option is "Pseudo unimolecular reaction"

In the reaction; $A + 2B \to 3C + D$ which of the following expressions does not describe changes in the concentration of various species as a function of time?

For the given reaction, $\frac{{ - d\left[ A \right]}}{{dt}} = - \frac{1}{2}\frac{{d\left[ B \right]}}{{dt}} = \frac{1}{3}\frac{{d\left[ C \right]}}{{dt}} = \frac{{d\left[ D \right]}}{{dt}}$

$\frac{{d\left[ C \right]}}{{dt}} = \frac{{ - 3d\left[ A \right]}}{{dt}}$

$\frac{{3d\left[ D \right]}}{{dt}} = \frac{{d\left[ C \right]}}{{dt}}$

$\frac{{3d\left[ B \right]}}{{dt}} = \frac{{ - 2d\left[ C \right]}}{{dt}}$

The correct Option is "$\frac{{2d\left[ B \right]}}{{dt}} = \frac{{d\left[ A \right]}}{{dt}}$"

The rate of disappearance of ${\rm{S}}{{\rm{O}}_2}$ in the reaction; $2{\rm{S}}{{\rm{O}}_2} + {{\rm{O}}_2} \to 2{\rm{S}}{{\rm{O}}_3}$ is $1.28 \times {10^{ - 3}}\;{\rm{g}}/{\rm{sec}}{\rm{.}}$ Then the rate of formation of ${\rm{S}}{{\rm{O}}_3}$ is:

Rate of formation of ${\rm{S}}{{\rm{O}}_3}$ = rate of disappearance of ${\rm{S}}{{\rm{O}}_2}$ $ = 1.28 \times {10^{ - 3}}\;{\rm{g}}/\sec = \frac{{1.28 \times {{10}^{ - 3}}}}{{64}}\;{\rm{M}}/\sec $ $ = \frac{{1.28 \times {{10}^{ - 3}}}}{{64}} \times 80\;{\rm{g}}/\sec = 1.60 \times {10^{ - 3}}\;{\rm{g}}/\sec $

$0.64 \times {10^{ - 3}}\;{\rm{g}}/\sec $

$0.80 \times {10^{ - 3}}\;{\rm{g}}/\sec $

$1.28 \times {10^{ - 1}}\;{\rm{g}}/\sec $

The correct Option is "$1.60 \times {10^{ - 3}}\;{\rm{g}}/\sec $"

In the reaction $2A + B \to {A_2}B,$ if the concentration of A is doubled and of B is halved, then the rate of the reaction will

$2A + B \to {A_2}B,$ ${r_1} = k{\left[ A \right]^2}\left[ B \right]$ ${r_2} = k{\left[ {2A} \right]^2}\left[ {B/2} \right]$ $Or\;\;{r_2} = 2k{\left[ A \right]^2}\left[ B \right]$ $\therefore \;\;{r_2} = 2{r_1}$

The correct Option is "Increase by two times"

Which of the following is the predominant product in the reaction of ${\rm{HOBr}}$ with propene?

The addition takes place according to Markownikoff’s rule.

$2 - {\rm{bromo}} - 2 - {\rm{propanol}}$

The correct Option is "1-bromo-2-propanol"

Which one of the following is an intermediate in the reaction of benzene with ${\rm{C}}{{\rm{H}}_3}{\rm{Cl}}$ in the presence of anhydrous ${\rm{AlC}}{{\rm{l}}_3}$?

${\rm{CH}}_3^ + $ acts as an intermediate in the given reaction (Friedel Craft’s alkylation). It is an example of electrophilic aromatic substitution. In this reaction ${\rm{CH}}_3^ + $ is electrophile.

The correct Option is "${\rm{CH}}_3^ + $"

The reagent ${\rm{X}}$ in the reactions

Oxymercuration-demercuration is an example of hydration of alkene according to Markownikoff’s rule

The correct Option is "${\left( {{\rm{C}}{{\rm{H}}_3}{\rm{COO}}} \right)_2}{\rm{Hg}}$"

Identify ‘A’ in the reaction:

${\rm{C}}{{\rm{H}}_3}{\rm{C}} \equiv {\rm{CH}} = {\rm{C}}{{\rm{H}}_2}$

Product [A] in the reaction :

In such cases where migrating group is cycloalkyl group, ring expansion may occur.

In the reaction; the products are:

Ether reacts with acid to give protonated ether. The next step involves nucleophilic attack by halide ion with the displacement of weakly basic alcohol molecule. .

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