AP EAMCET 2021 Chemistry Question Paper with Answer and Solution

(A) The momentum of a photon is given by $p = \frac{h}{\lambda}$.
For an electron,the momentum is $p = mv$.
Equating the two,we get $mv = \frac{h}{\lambda}$,which implies $v = \frac{h}{m \lambda}$.
Given:
$h = 6.63 \times 10^{-34} \ J \cdot s$
$m = 9.1 \times 10^{-31} \ kg$
$\lambda = 663 \ nm = 663 \times 10^{-9} \ m = 6.63 \times 10^{-7} \ m$
Substituting the values:
$v = \frac{6.63 \times 10^{-34}}{(9.1 \times 10^{-31}) \times (6.63 \times 10^{-7})}$
$v = \frac{10^{-34}}{9.1 \times 10^{-31} \times 10^{-7}} = \frac{10^{-34}}{9.1 \times 10^{-38}}$
$v = \frac{10^4}{9.1} \approx 1098.9 \ m/s$.
Thus,the velocity is approximately $1098 \ m/s$.

(D) Given: Precision $= \pm 1 \%$,$\Delta x = 1.05 \times 10^{-13} \ m$,$h = 6.6 \times 10^{-34} \ kg \ m^2 \ s^{-1}$.
Velocity of light $c = 3 \times 10^8 \ m/s$.
Particle velocity $v = 2c = 6 \times 10^8 \ m/s$.
Uncertainty in velocity $\Delta v = 1 \% \text{ of } v = 0.01 \times 6 \times 10^8 = 6 \times 10^6 \ m/s$.
According to Heisenberg's Uncertainty Principle: $\Delta x \cdot m \cdot \Delta v \ge \frac{h}{4 \pi}$.
$m = \frac{h}{4 \pi \cdot \Delta x \cdot \Delta v}$.
$m = \frac{6.6 \times 10^{-34}}{4 \times 3.14 \times 1.05 \times 10^{-13} \times 6 \times 10^6}$.
$m = \frac{6.6 \times 10^{-34}}{79.128 \times 10^{-7}} \approx 0.0834 \times 10^{-27} \ kg = 0.834 \times 10^{-28} \ kg$.
Rounding to the nearest provided option,the correct value is $0.8 \times 10^{-28} \ kg$.

(B) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta P \geq \frac{h}{4 \pi}$.
Given that the uncertainty in momentum $(\Delta P)$ is equal to the uncertainty in position $(\Delta x)$,i.e.,$\Delta P = \Delta x$.
Substituting $\Delta x = \Delta P$ in the uncertainty principle equation:
$(\Delta P)^2 \geq \frac{h}{4 \pi}$.
Since $\Delta P = m \cdot \Delta v$,we have $(m \cdot \Delta v)^2 \geq \frac{h}{4 \pi}$.
$m^2 \cdot (\Delta v)^2 \geq \frac{h}{4 \pi}$.
$(\Delta v)^2 \geq \frac{h}{4 \pi m^2}$.
Taking the square root on both sides:
$\Delta v \geq \sqrt{\frac{h}{4 \pi m^2}} = \frac{1}{2 m} \sqrt{\frac{h}{\pi}}$.

(D) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$,where $\Delta p = m \Delta v$.
Thus,$\Delta x \geq \frac{h}{4 \pi m \Delta v}$.
Given velocity $v = 2.99 \times 10^4 \ cm \ s^{-1}$ and accuracy $0.0016 \%$.
Uncertainty in velocity $\Delta v = v \times \frac{0.0016}{100} = 2.99 \times 10^4 \times 1.6 \times 10^{-5} = 0.4784 \ cm \ s^{-1}$.
Substituting the values:
$\Delta x = \frac{6.626 \times 10^{-27}}{4 \times 3.1416 \times 9.1 \times 10^{-28} \times 0.4784}$.
$\Delta x = \frac{6.626 \times 10^{-27}}{5.475 \times 10^{-27}} \approx 1.21 \ cm$.
Converting to $mm$: $1.21 \ cm = 12.1 \ mm$.

(D) The uncertainty in velocity $\Delta v$ is given by $0.5 \%$ of $3 \times 10^7 \text{ ms}^{-1}$.
$\Delta v = \frac{0.5}{100} \times 3 \times 10^7 \text{ ms}^{-1} = 1.5 \times 10^5 \text{ ms}^{-1}$.
According to Heisenberg's uncertainty principle,$\Delta x \times \Delta p \ge \frac{h}{4 \pi}$.
Since $\Delta p = m \Delta v$,we have $\Delta x = \frac{h}{4 \pi m \Delta v}$.
Substituting the values: $\Delta x = \frac{6.6 \times 10^{-34}}{4 \times 3.14 \times 1.66 \times 10^{-27} \times 1.5 \times 10^5}$.
$\Delta x = \frac{6.6 \times 10^{-34}}{31.27 \times 10^{-22}} \approx 2.11 \times 10^{-13} \text{ m}$.

(C) According to the photoelectric effect equation:
$h\nu = KE_{max} + h\nu_0$
Where $h$ is Planck's constant $(6.63 \times 10^{-34} \ J \cdot s)$,$\nu$ is the incident frequency,$KE_{max}$ is the maximum kinetic energy,and $\nu_0$ is the threshold frequency.
Rearranging for $\nu_0$:
$h\nu_0 = h\nu - KE_{max}$
$\nu_0 = \nu - \frac{KE_{max}}{h}$
Substituting the given values:
$\nu_0 = (4 \times 10^{14} \ s^{-1}) - \frac{6.63 \times 10^{-20} \ J}{6.63 \times 10^{-34} \ J \cdot s}$
$\nu_0 = 4 \times 10^{14} \ s^{-1} - 1 \times 10^{14} \ s^{-1}$
$\nu_0 = 3 \times 10^{14} \ s^{-1}$

(C) According to de-Broglie's wavelength,$\lambda = \frac{h}{mv}$.
For particle $A$,$\lambda_A = \frac{h}{m_A v_A}$.
For particle $B$,$\lambda_B = \frac{h}{m_B v_B}$.
The ratio of wavelengths is given as $\frac{\lambda_A}{\lambda_B} = \frac{2}{1}$.
Substituting the expressions,we get $\frac{\lambda_A}{\lambda_B} = \frac{m_B v_B}{m_A v_A} = 2$.
Given $v_A = 0.05 \ ms^{-1}$ and $v_B = 0.02 \ ms^{-1}$,we have $\frac{m_B \times 0.02}{m_A \times 0.05} = 2$.
$\frac{m_B}{m_A} \times \frac{2}{5} = 2$.
$\frac{m_B}{m_A} = 2 \times \frac{5}{2} = 5$.
Therefore,$\frac{m_A}{m_B} = \frac{1}{5}$,which is $1: 5$.

(B) According to the de-Broglie hypothesis for a stationary orbit,the circumference of the orbit is an integral multiple of the wavelength:
$n \lambda = 2 \pi r$
$\lambda = \frac{2 \pi r}{n}$
For a hydrogen atom,the radius of the $n^{th}$ orbit is given by $r = a_0 \times n^2$.
Substituting this into the wavelength formula:
$\lambda = \frac{2 \pi (a_0 \times n^2)}{n} = 2 \pi a_0 n$
Given $n = 2$ and $a_0 = 0.529 \ \mathring{A}$:
$\lambda = 2 \times \pi \times 0.529 \times 2 \ \mathring{A}$
$\lambda = 2.116 \pi \ \mathring{A}$

(A) The azimuthal quantum number $(l)$ determines the subshell type. For a given subshell,the number of orbitals is given by $(2l + 1)$.
Each orbital can hold a maximum of $2$ electrons.
Therefore,the maximum number of electrons in a subshell is $2 \times (2l + 1)$.
Given $n=3$ and $l=2$,this corresponds to the $3d$ subshell.
Substituting $l=2$ into the formula: $2 \times (2(2) + 1) = 2 \times (4 + 1) = 2 \times 5 = 10$ electrons.
Thus,the $3d$ subshell can accommodate a maximum of $10$ electrons.

(B) The number of angular nodes is given by the azimuthal quantum number $l$. For $d$-orbitals,$l = 2$,so they have $2$ angular nodes.
The number of radial nodes is calculated using the formula: $\text{Radial nodes} = n - l - 1$.
For the $3d$ orbital: $n = 3$ and $l = 2$.
$\text{Radial nodes} = 3 - 2 - 1 = 0$.
Thus,the $3d$ orbital has $0$ radial nodes and $2$ angular nodes.

(A) The angular momentum of an electron in an orbital is given by the formula: $\text{Angular momentum} = \sqrt{l(l+1)} \frac{h}{2\pi}$.
If the angular momentum is $0$,then $\sqrt{l(l+1)} \frac{h}{2\pi} = 0$,which implies $l(l+1) = 0$.
This means the azimuthal quantum number $l$ must be equal to $0$.
The $s$-orbital corresponds to $l = 0$.
Therefore,the $3s$ orbital is the correct answer.

(A) The electronic configurations given in options $B$,$C$,and $D$ are $[ Ar ] 3 d^{10} 4 s^2$,$[ Kr ] 4 d^{10} 5 s^2$,and $[ Xe ] 4 f^{14} 5 d^{10} 6 s^2$ respectively.
These configurations correspond to the general valence shell configuration of $(n-1)d^{10} ns^2$,which is characteristic of Group $12$ elements (Zinc,Cadmium,and Mercury).
Option $A$ has the configuration $[ Ne ] 3 s^2 3 p^5$,which corresponds to Chlorine (Group $17$,Halogen family).
Therefore,the element in option $A$ does not belong to the same family as the others.

(B) The given electronic configuration is $(n-1)d^2 ns^2$ with $n=4$.
Since the principal quantum number $n$ represents the period number,the element belongs to the $4^{\text{th}}$ period.
For $d$-block elements,the group number is calculated as the sum of electrons in the $(n-1)d$ subshell and the $ns$ subshell.
Group number = (Number of electrons in $(n-1)d$) + (Number of electrons in $ns$) = $2 + 2 = 4$.
Therefore,the element belongs to the $4^{\text{th}}$ period and $4^{\text{th}}$ group.

(D) The number of electrons in a species is calculated by subtracting the charge from the atomic number of the neutral atom.
$1$. For $Be^{2+}$: Atomic number of $Be$ is $4$. Electrons = $4 - 2 = 2$. Thus,$(i-c)$.
$2$. For $H^{+}$: Atomic number of $H$ is $1$. Electrons = $1 - 1 = 0$. Thus,$(ii-a)$.
$3$. For $Na^{+}$: Atomic number of $Na$ is $11$. Electrons = $11 - 1 = 10$. Thus,$(iii-b)$.
$4$. For $Mg^{+}$: Atomic number of $Mg$ is $12$. Electrons = $12 - 1 = 11$. Thus,$(iv-d)$.
Therefore,the correct match is $(i-c), (ii-a), (iii-b), (iv-d)$.

(B) The industrial preparation of methanol involves the catalytic hydrogenation of carbon monoxide.
The chemical reaction is: $CO(g) + 2H_2(g) \xrightarrow[ZnO-Cr_2O_3]{573-673 \ K, 200-300 \ atm} CH_3OH(g)$.
Comparing this with the given ratio of $1:2$ for gases $X$ and $Y$,we identify $X = CO$ and $Y = H_2$.

(D) Extensive properties are those that depend on the amount of matter or mass present in the system.
$I$. Molar entropy is an intensive property because it is defined per mole.
$II$. Specific volume is an intensive property because it is defined per unit mass.
$III$. Heat capacity is an extensive property as it depends on the total mass of the substance.
$IV$. Volume is an extensive property as it depends on the total amount of matter.
Therefore,$III$ and $IV$ are extensive properties.

(B) For an ideal gas,the equation of state is $PV = nRT$.
Since the pressure $P$ is constant,the change in volume $\Delta V$ due to a change in temperature $\Delta T$ is given by $P \Delta V = nR \Delta T$.
The work done $W$ in a process at constant pressure is defined as $W = P \Delta V$.
Substituting the expression for $P \Delta V$,we get $W = nR \Delta T$.
Given $n = 2$ moles and $\Delta T = 20 \ K$ (or $20^{\circ} C$),the work done is $W = 2 \times R \times 20 = 40R$.

(B) The total kinetic energy $(K.E.)$ of an ideal gas is given by the formula: $K.E. = \frac{3}{2} nRT$.
For flask $A$ containing $H_2$: $n_A = \frac{m}{M_{H_2}} = \frac{m}{2}$ and $T_A = 300 \ K$.
Thus,$(K.E.)_A = \frac{3}{2} \times \frac{m}{2} \times R \times 300$.
For flask $B$ containing $CO_2$: $n_B = \frac{m}{M_{CO_2}} = \frac{m}{44}$ and $T_B = 600 \ K$.
Thus,$(K.E.)_B = \frac{3}{2} \times \frac{m}{44} \times R \times 600$.
Taking the ratio: $\frac{(K.E.)_A}{(K.E.)_B} = \frac{\frac{m}{2} \times 300}{\frac{m}{44} \times 600} = \frac{300/2}{600/44} = \frac{150}{600/44} = \frac{150 \times 44}{600} = \frac{44}{4} = 11:1$.

(B) The pressure $P = 10^7 \ N \cdot m^{-2}$.
Initial volume $V_i = 5 \ m^3$.
Final volume $V_f = 10 \ m^3$.
Change in volume $\Delta V = V_f - V_i = 10 \ m^3 - 5 \ m^3 = 5 \ m^3$.
The work done on the gas during expansion against constant external pressure is given by $W = -P_{ext} \Delta V$.
Substituting the values: $W = -(10^7 \ N \cdot m^{-2}) \times (5 \ m^3) = -50 \times 10^6 \ J$.
Since $1 \ MJ = 10^6 \ J$,we get $W = -50 \ MJ$.

(B) For an isothermal process,the temperature remains constant,so $\Delta T = 0$.
Since the internal energy $(U)$ and enthalpy $(H)$ of an ideal gas are functions of temperature only,$\Delta T = 0$ implies $\Delta U = 0$ and $\Delta H = 0$.
According to the first law of thermodynamics,$\Delta U = Q + W$.
Substituting $\Delta U = 0$,we get $0 = Q + W$,which means $W = -Q$.
Therefore,statements $(i)$,$(ii)$,and $(iv)$ are true,while $(iii)$ is false.

(C) Helium $(He)$ and Neon $(Ne)$ are both monoatomic gases.
For a monoatomic gas,the degree of freedom $(f)$ is $3$.
The ratio of molar specific heat capacities $\gamma = \frac{C_p}{C_V} = 1 + \frac{2}{f}$.
Since both gases are monoatomic,the mixture will also behave as a monoatomic gas with $f = 3$.
Therefore,$\frac{C_p}{C_V} = 1 + \frac{2}{3} = \frac{5}{3}$.

(D) For an irreversible expansion of a gas against a constant external pressure,the work done is given by: $W = -p_{\text{ext}} \Delta V$.
Given $p_{\text{ext}} = 10 \ bar$,$V_1 = 20 \ L$,and $V_2 = 30 \ L$.
$W = -10 \ bar \times (30 \ L - 20 \ L) = -100 \ L \ bar$.
Since $1 \ L \ bar = 100 \ J$,$W = -100 \times 100 \ J = -10000 \ J = -10 \ kJ$.
For an isoenthalpic process,$\Delta H = 0$.
We know $\Delta H = \Delta U + \Delta(PV) = 0$.
From the first law of thermodynamics,$\Delta U = Q + W$,so $Q + W + \Delta(PV) = 0$.
For an ideal gas,$\Delta H = nC_p\Delta T = 0$,which implies $\Delta T = 0$.
Since $\Delta U = nC_v\Delta T$,$\Delta U = 0$.
Thus,$0 = Q + W$,which means $Q = -W$.
$Q = -(-10 \ kJ) = 10 \ kJ$.

(D) According to the first law of thermodynamics,the change in internal energy $(\Delta U)$ of a system is given by the equation: $\Delta U = Q - W$,where $Q$ is the heat exchanged and $W$ is the work done by the system.
If the process is adiabatic,there is no exchange of heat,so $Q = 0$.
Substituting this into the equation gives $\Delta U = -W$.
This implies that the change in internal energy is equal to the work done on the system (or the negative of the work done by the system) during an adiabatic process.
Therefore,the change in internal energy equals adiabatic work.

(C) The reaction for the dissociation is: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
At $t=0$,we have $1 \ mol$ of $N_2O_4$ and $0 \ mol$ of $NO_2$.
At equilibrium,with $50 \%$ dissociation,we have $1-0.5 = 0.5 \ mol$ of $N_2O_4$ and $2 \times 0.5 = 1 \ mol$ of $NO_2$.
The total number of moles is $0.5 + 1 = 1.5 \ mol$.
The partial pressures are:
$P_{N_2O_4} = \frac{0.5}{1.5} \times 1 \ atm = \frac{1}{3} \ atm$
$P_{NO_2} = \frac{1}{1.5} \times 1 \ atm = \frac{2}{3} \ atm$
The equilibrium constant $K_p$ is:
$K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(2/3)^2}{1/3} = \frac{4/9}{1/3} = \frac{4}{3} \approx 1.33 \ atm$
The standard free energy change is given by:
$\Delta G^{\circ} = -RT \ln K_p = -2.303 \times RT \times \log_{10} K_p$
$\Delta G^{\circ} = -2.303 \times (8.314 \ J \cdot K^{-1} \cdot mol^{-1}) \times (333 \ K) \times \log_{10} 1.33$
$\Delta G^{\circ} = -2.303 \times 8.314 \times 333 \times 0.1239 \approx -790 \ J \cdot mol^{-1}$

(B) $C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$
$\Delta n_{g} = \text{Number of gaseous products} - \text{number of gaseous reactants}$
$\Delta n_{g} = 1 - \frac{1}{2} = 0.5$
$R = 8.314 \ J \cdot K^{-1} \ mol^{-1}$
$T = 298 \ K$
$\Delta H = \Delta U + \Delta n_{g} RT$
$\Delta H - \Delta U = \Delta n_{g} RT$
$\Delta H - \Delta U = 0.5 \times 8.314 \times 298 = 1238.786 \ J \cdot mol^{-1}$
$\text{The approximate value is } 1238 \ J \cdot mol^{-1}$.

(A) The enthalpy of neutralisation for a strong acid and strong base is given by: $H^{+} + OH^{-} \longrightarrow H_2O; \Delta_r H^{\circ} = -55.84 \ kJ \ mol^{-1}$.
For the weak acid $CH_3COOH$,the neutralisation reaction is: $CH_3COOH + OH^{-} \longrightarrow CH_3COO^{-} + H_2O; \Delta_r H = -49.86 \ kJ \ mol^{-1}$.
The enthalpy of ionisation $(\Delta H_i)$ is the energy required to dissociate the weak acid: $CH_3COOH \longrightarrow CH_3COO^{-} + H^{+}; \Delta H_i = ?$.
This can be calculated as: $\Delta H_i = \Delta H_{\text{neutralisation(weak)}} - \Delta H_{\text{neutralisation(strong)}}$.
$\Delta H_i = -49.86 - (-55.84) = 5.98 \ kJ \ mol^{-1}$.

(C) Molar mass of benzoic acid $(C_6H_5COOH) = 122 \ g \ mol^{-1}$.
Heat gained by water $(Q) = m \times s \times \Delta T$.
$Q = 18.94 \times 10^3 \ g \times 0.998 \ cal \ g^{-1} \ ^{\circ}C^{-1} \times 0.632 \ ^{\circ}C$.
$Q = 11946.3 \ cal = 11.946 \ kcal = 49.98 \ kJ$.
Heat liberated by $1.89 \ g$ of benzoic acid $= 49.98 \ kJ$.
Heat of combustion for $1 \ mol$ $(122 \ g)$ $= \frac{49.98 \ kJ}{1.89 \ g} \times 122 \ g \ mol^{-1} \approx 3226 \ kJ \ mol^{-1}$.
The closest option is $3240 \ kJ \ mol^{-1}$.

(A) The process of vaporization is non-spontaneous below the boiling point at $1 \ atm$ pressure.
For any process,the spontaneity is determined by the Gibbs free energy change,$\Delta G = \Delta H - T \Delta S$.
At the boiling point $(T_{b} = 373 \ K)$,the system is at equilibrium,so $\Delta G = 0$,which implies $\Delta S = \Delta H / T_{b}$.
For vaporization,$\Delta H > 0$ and $\Delta S > 0$.
At $T = 75^{\circ} C = 348 \ K$,since $T < T_{b}$,the term $T \Delta S$ is less than $\Delta H$.
Therefore,$\Delta G = \Delta H - T \Delta S > 0$.
Since $\Delta G > 0$,the process is non-spontaneous at $75^{\circ} C$.

(B) The reaction is $\frac{1}{2} X_2 + \frac{3}{2} Y_2 \rightarrow XY_3$ with $\Delta H = -30 \ kJ \ mol^{-1} = -30000 \ J \ mol^{-1}$.
First,calculate the entropy change of the reaction $\Delta S^{\circ}$:
$\Delta S^{\circ} = S^{\circ}(XY_3) - [\frac{1}{2} S^{\circ}(X_2) + \frac{3}{2} S^{\circ}(Y_2)]$
$\Delta S^{\circ} = 50 - [\frac{1}{2} \times 60 + \frac{3}{2} \times 40]$
$\Delta S^{\circ} = 50 - [30 + 60] = 50 - 90 = -40 \ J \ K^{-1} \ mol^{-1}$.
At equilibrium,$\Delta G = 0$,so $\Delta H = T \Delta S$.
$T = \frac{\Delta H}{\Delta S} = \frac{-30000 \ J \ mol^{-1}}{-40 \ J \ K^{-1} \ mol^{-1}} = 750 \ K$.

(C) The spontaneity of a reaction is determined by the Gibbs free energy equation: $\Delta G = \Delta H - T \Delta S$.
For a reaction to be spontaneous,$\Delta G < 0$.
For a reaction to be non-spontaneous,$\Delta G > 0$.
Given that the reaction is non-spontaneous at $298 \ K$ $(\Delta G > 0)$ and spontaneous at $350 \ K$ $(\Delta G < 0)$,the reaction must be endothermic $(\Delta H > 0)$ and have a positive entropy change $(\Delta S > 0)$.
At lower temperatures,the $T \Delta S$ term is smaller than $\Delta H$,making $\Delta G$ positive.
At higher temperatures,the $T \Delta S$ term exceeds $\Delta H$,making $\Delta G$ negative.
Thus,the conditions are $\Delta H > 0$ and $\Delta S > 0$.

(B) The given reaction is the vaporization of water at its boiling point: $H_2O_{(l)} \longrightarrow H_2O_{(g)}$.
This process occurs at equilibrium at $T=100^{\circ} C$ and $P=1 \ atm$.
For any process at equilibrium,the change in entropy of the universe is zero: $\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} = 0$.
Therefore,$\Delta S_{\text{system}} = -\Delta S_{\text{surroundings}}$.
Since vaporization involves a transition from a liquid to a gas,the disorder increases,so $\Delta S_{\text{system}} > 0$.
Consequently,$\Delta S_{\text{surroundings}}$ must be negative,i.e.,$\Delta S_{\text{surroundings}} < 0$.

(D) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^{\circ} = -RT \ln K$.
At standard states,the reaction quotient $Q$ is equal to $1$.
At equilibrium,$\Delta G = 0$.
However,the question specifies standard states,implying $\Delta G^{\circ} = -RT \ln K$.
For the reaction to be at equilibrium under standard conditions,$\Delta G^{\circ}$ must be $0$.
Therefore,$0 = -RT \ln K$,which implies $\ln K = 0$.
Thus,$K = e^0 = 1$.

(A) The reaction is the acidic cleavage of a cyclic ether with $HI$. The ether oxygen gets protonated by $H^+$. The cleavage occurs at the bond that leads to the formation of a more stable carbocation. In this case,the bond between the oxygen and the tertiary carbon atom breaks to form a stable benzylic-tertiary carbocation. The iodide ion $(I^-)$ then attacks this carbocation to form the final product. The structure of the product is $2-(2-iodo-propan-2-yl)phenol$ or a related derivative where the ring is opened,specifically forming an alcohol and an alkyl iodide. The correct product is the one where the $I$ is attached to the tertiary carbon and the $OH$ is attached to the primary carbon of the side chain.

(D) The reaction is an esterification reaction between salicylic acid $(2-hydroxybenzoic acid)$ and methanol $(MeOH)$ in the presence of a concentrated acid catalyst $(H_2SO_4)$.
In this reaction,the carboxylic acid group $(-COOH)$ reacts with the alcohol $(-OH)$ to form an ester $(-COOCH_3)$.
The phenolic $-OH$ group is less reactive towards esterification under these conditions compared to the carboxylic acid group.
Therefore,the major product is methyl salicylate,where the carboxylic acid group is converted to a methyl ester.

(D) The correct answer is $D$.
Molecularity is a theoretical concept defined as the number of reacting species (atoms,ions,or molecules) taking part in an elementary reaction,which must collide simultaneously in order to bring about a chemical reaction.
It is determined by examining the balanced chemical equation of an elementary step.
In contrast,the order of a reaction is an experimental quantity determined from the rate law.

(D) For the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$,the rate of reaction is expressed as:
$Rate = -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$
Equating the terms for $NH_3$ and $H_2$:
$-\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$
Multiplying both sides by $6$:
$-2 \frac{d[H_2]}{dt} = 3 \frac{d[NH_3]}{dt}$
Rearranging gives $3 \frac{d[NH_3]}{dt} = -2 \frac{d[H_2]}{dt}$.

(C) For the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$,the rate of reaction is expressed as:
Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
Given that $-\frac{d[N_2]}{dt} = 0.02 \ mol \ L^{-1} \ s^{-1}$.
Equating the terms for $N_2$ and $H_2$:
$-\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt}$.
Therefore,$-\frac{d[H_2]}{dt} = 3 \times (-\frac{d[N_2]}{dt})$.
$-\frac{d[H_2]}{dt} = 3 \times 0.02 \ mol \ L^{-1} \ s^{-1} = 0.06 \ mol \ L^{-1} \ s^{-1}$.

(C) Since the given three lines are concurrent,the determinant of their coefficients must be zero:
$\left|\begin{array}{rrr} 4 & 3 & -1 \\ 1 & -1 & 5 \\ k & 5 & -3 \end{array}\right| = 0$
Expanding the determinant along the first row:
$4((-1)(-3) - (5)(5)) - 3((1)(-3) - (5)(k)) - 1((1)(5) - (-1)(k)) = 0$
$4(3 - 25) - 3(-3 - 5k) - 1(5 + k) = 0$
$4(-22) + 9 + 15k - 5 - k = 0$
$-88 + 9 - 5 + 14k = 0$
$-84 + 14k = 0$
$14k = 84$
$k = 6$

(D) The given equation of the pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$.
Comparing this with the general form $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=1, b=-35, g=-2, f=22, c=-12$.
The point of intersection $(x_0, y_0)$ of the pair of lines is given by the formula:
$x_0 = \frac{hf-bg}{ab-h^2} = \frac{(1)(22)-(-35)(-2)}{(1)(-35)-(1)^2} = \frac{22-70}{-35-1} = \frac{-48}{-36} = \frac{4}{3}$
$y_0 = \frac{gh-af}{ab-h^2} = \frac{(-2)(1)-(1)(22)}{(1)(-35)-(1)^2} = \frac{-2-22}{-36} = \frac{-24}{-36} = \frac{2}{3}$
Since the lines are concurrent,the point $(\frac{4}{3}, \frac{2}{3})$ must satisfy the line $5x+\lambda y-8=0$.
$5(\frac{4}{3}) + \lambda(\frac{2}{3}) - 8 = 0$
$\frac{20}{3} + \frac{2\lambda}{3} - 8 = 0$
Multiply by $3$: $20 + 2\lambda - 24 = 0$
$2\lambda - 4 = 0$
$2\lambda = 4$
$\lambda = 2$

(B) The areal velocity $A$ is defined as the rate at which the area is swept by the position vector of the planet.
$A = \frac{dA}{dt} = \frac{1}{2} r^2 \omega$
Multiplying both sides by the mass $M$ of the planet:
$M A = \frac{1}{2} M r^2 \omega$
Since the moment of inertia $I = M r^2$,we can substitute this into the equation:
$M A = \frac{1}{2} I \omega$
We know that the angular momentum $L$ is given by $L = I \omega$.
Therefore,$M A = \frac{1}{2} L$.
Rearranging for $L$,we get:
$L = 2 M A$.

(C) Step $1$: Dehydrohalogenation of $C_6H_5-CHBr-CH_2Br$ with $alc. KOH$ followed by $NaNH_2$ leads to the formation of phenylacetylene $(C_6H_5-C \equiv CH)$.
Step $2$: The cyclic trimerization of phenylacetylene occurs when passed through a red hot iron tube at $873 \ K$.
Step $3$: The trimerization of $C_6H_5-C \equiv CH$ yields $1,3,5-triphenylbenzene$ as the major product due to steric hindrance,which favors the meta-substituted product.