Rate of a reaction Questions in English

(C) Given reaction: $A_2B_{3(g)} \to A_{2(g)} + \frac{3}{2}B_{2(g)}$
Initial pressure $(t=0)$: $P_0 = 60 \ torr$
Pressure at $t = 2.5 \ min$:
$P(A_2B_3) = 60 - x$
$P(A_2) = x$
$P(B_2) = \frac{3}{2}x$
Total pressure $P_t = (60 - x) + x + \frac{3}{2}x = 60 + \frac{3}{2}x$
$75 = 60 + 1.5x$ $\Rightarrow 1.5x = 15$ $\Rightarrow x = 10 \ torr$
Rate of disappearance of $A_2B_3 = \frac{x}{t} = \frac{10}{2.5} = 4 \ torr \cdot min^{-1}$.

(C) The rate law for the reaction $xA \to yB$ is given by: $\frac{1}{x} (-\frac{d[A]}{dt}) = \frac{1}{y} (+\frac{d[B]}{dt})$.
Given equation: $0.48 + \log \{-\frac{d[A]}{dt}\} = \log \{+\frac{d[B]}{dt}\} + 0.7$.
Using $\log 3 \approx 0.477 \approx 0.48$ and $\log 5 \approx 0.699 \approx 0.7$,we rewrite the equation:
$\log 3 + \log \{-\frac{d[A]}{dt}\} = \log \{+\frac{d[B]}{dt}\} + \log 5$.
$\log \{3 \times (-\frac{d[A]}{dt})\} = \log \{5 \times (+\frac{d[B]}{dt})\}$.
$3 (-\frac{d[A]}{dt}) = 5 (+\frac{d[B]}{dt})$.
Rearranging gives: $\frac{-\frac{d[A]}{dt}}{5} = \frac{+\frac{d[B]}{dt}}{3}$.
Comparing this with the stoichiometric rate expression $\frac{-\frac{d[A]}{dt}}{x} = \frac{+\frac{d[B]}{dt}}{y}$,we get $x = 5$ and $y = 3$.
Therefore,$x : y = 5 : 3$.

(B) The rate of reaction in terms of pressure is given by $\frac{dp}{dt} = \frac{P_1 - P_2}{\Delta t} = \frac{2 - 1.2}{50 \times 60} \, atm \, s^{-1} = \frac{0.8}{3000} \, atm \, s^{-1} = 2.66 \times 10^{-4} \, atm \, s^{-1}$.
Using the ideal gas equation $PV = nRT$,we have $P = CRT$,so $\frac{dP}{dt} = RT \frac{dC}{dt}$.
Here,$R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$ and $T = 27 + 273 = 300 \, K$.
Therefore,$\frac{dC}{dt} = \frac{1}{RT} \times \frac{dP}{dt} = \frac{2.66 \times 10^{-4}}{0.0821 \times 300} \, M s^{-1}$.
$\frac{dC}{dt} = \frac{2.66 \times 10^{-4}}{24.63} \approx 1.08 \times 10^{-5} \, M s^{-1}$.

(C) For a general reaction $aA + bB \to cC + dD$,the rate of reaction is expressed as:
Rate $= -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = +\frac{1}{c}\frac{d[C]}{dt} = +\frac{1}{d}\frac{d[D]}{dt}$
Given expression:
$+\frac{1}{2}\frac{d[C]}{dt} = -\frac{1}{3}\frac{d[D]}{dt} = +\frac{1}{4}\frac{d[A]}{dt} = -\frac{d[B]}{dt}$
Comparing the coefficients:
For $C$,the coefficient is $2$.
For $D$,the coefficient is $3$ (negative sign indicates reactant).
For $A$,the coefficient is $4$.
For $B$,the coefficient is $1$ (negative sign indicates reactant).
Thus,the reactants are $B$ and $D$ with stoichiometric coefficients $1$ and $3$ respectively,and the products are $A$ and $C$ with stoichiometric coefficients $4$ and $2$ respectively.
The balanced chemical equation is: $B + 3D \to 4A + 2C$.

(D) For the reaction $2SO_2 + O_2 \to 2SO_3$,the rate expression is given by:
$-\frac{1}{2} \frac{d[SO_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt}$.
Since the stoichiometric coefficients of $SO_2$ and $SO_3$ are both $2$,the rate of disappearance of $SO_2$ is equal to the rate of appearance of $SO_3$.
Therefore,the rate of appearance of $SO_3 = 1.28 \times 10^{-5} \ mol \ s^{-1}$.

(A) The stoichiometric equation is $2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$.
According to the rate law,the rate of reaction is given by: $-\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{d[O_2]}{dt}$.
Therefore,the rate of disappearance of $N_2O_5$ is $-\frac{d[N_2O_5]}{dt} = 2 \times \frac{d[O_2]}{dt}$.
Given the rate of formation of $O_2$ is $0.032 \ g \ h^{-1}$,we convert this to molar rate: $\text{Rate in mol } h^{-1} = \frac{0.032 \ g \ h^{-1}}{32 \ g \ mol^{-1}} = 0.001 \ mol \ h^{-1}$.
The rate of disappearance of $N_2O_5$ in $mol \ h^{-1}$ is $2 \times 0.001 = 0.002 \ mol \ h^{-1}$.
Converting this back to mass rate: $0.002 \ mol \ h^{-1} \times 108 \ g \ mol^{-1} = 0.216 \ g \ h^{-1}$,which is approximately $0.22 \ g \ h^{-1}$.

(C) The rate of reaction $r$ for a non-zero order reaction depends on the concentration of reactants. Assuming a simple second-order reaction $r = k[A][B]$,we calculate the molar concentrations $(M = mol/L)$:
$I$. $[A] = 1 \ mol / 1 \ L = 1 \ M$,$[B] = 2 \ mol / 1 \ L = 2 \ M$. Rate $r_I = k(1)(2) = 2k$.
$II$. $[A] = 2 \ mol / 2 \ L = 1 \ M$,$[B] = 2 \ mol / 2 \ L = 1 \ M$. Rate $r_{II} = k(1)(1) = 1k$.
$III$. $[A] = 0.2 \ mol / 0.1 \ L = 2 \ M$,$[B] = 0.2 \ mol / 0.1 \ L = 2 \ M$. Rate $r_{III} = k(2)(2) = 4k$.
Comparing the rates,$r_{III} > r_I > r_{II}$. Therefore,sample $III$ reacts at the highest rate.

(D) For the reaction $2MnO_4^- + 10I^{-} + 16H^{+} \to 2Mn^{2+} + 5I_2 + 8H_2O$,the rate expression is given by:
Rate $= -\frac{1}{2} \frac{d[MnO_4^-]}{dt} = \frac{1}{5} \frac{d[I_2]}{dt}$
Given that the rate of disappearance of $MnO_4^-$ is $-\frac{d[MnO_4^-]}{dt} = 4.56 \times 10^{-3} \ M/s$.
Substituting this into the rate equation:
$\frac{1}{2} \times (4.56 \times 10^{-3}) = \frac{1}{5} \times \text{rate of appearance of } I_2$
Rate of appearance of $I_2 = \frac{5}{2} \times 4.56 \times 10^{-3} \ M/s$
Rate of appearance of $I_2 = 2.5 \times 4.56 \times 10^{-3} \ M/s = 1.14 \times 10^{-2} \ M/s$.

(B) For a general chemical reaction $aA + bB \to cC + dD$,the rate of reaction is given by:
Rate $= -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$.
For the given reaction $A + 2B \to 3C$,the stoichiometric coefficients are $a=1$,$b=2$,and $c=3$.
Substituting these values,we get:
Rate $= -\frac{d[A]}{dt} = -\frac{1}{2} \frac{d[B]}{dt} = \frac{1}{3} \frac{d[C]}{dt}$.

(D) For the reaction $2SO_2 + O_2 \rightleftharpoons 2SO_3$,the rate of reaction is expressed as:
Rate $= -\frac{1}{2} \frac{d[SO_2]}{dt} = -\frac{d[O_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt}$
Given that $\frac{d[O_2]}{dt} = -2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
Equating the terms for $SO_2$ and $O_2$:
$-\frac{1}{2} \frac{d[SO_2]}{dt} = -\frac{d[O_2]}{dt}$
$\frac{d[SO_2]}{dt} = 2 \times \frac{d[O_2]}{dt}$
$\frac{d[SO_2]}{dt} = 2 \times (-2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1})$
$\frac{d[SO_2]}{dt} = -5.00 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.

(A) For a general reaction $aA + bB \to cC + dD$,the rate of reaction is expressed as:
Rate $= -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt} = k[A]^n[B]^m$
Given the reaction $3A + 2B \to C + D$,the rate expression is:
Rate $= -\frac{1}{3} \frac{d[A]}{dt} = -\frac{1}{2} \frac{d[B]}{dt} = \frac{d[C]}{dt} = \frac{d[D]}{dt} = k[A]^n[B]^m$
Comparing this with the given options,the correct representation is $-\frac{1}{3} \frac{d[A]}{dt} = \frac{d[C]}{dt} = k[A]^n[B]^m$.

(B) Given: The rate of disappearance of $MnO_4^-$ is $-\frac{d[MnO_4^-]}{dt} = 4.56 \times 10^{-3} \ M s^{-1}$.
From the stoichiometry of the reaction $2MnO_4^- + 10I^{-} + 16H^{+} \to 2Mn^{2+} + 5I_2 + 8H_2O$,the rate expression is:
$-\frac{1}{2} \frac{d[MnO_4^-]}{dt} = \frac{1}{5} \frac{d[I_2]}{dt}$.
Rearranging to find the rate of appearance of $I_2$:
$\frac{d[I_2]}{dt} = -\frac{5}{2} \frac{d[MnO_4^-]}{dt}$.
Substituting the given value:
$\frac{d[I_2]}{dt} = \frac{5}{2} \times (4.56 \times 10^{-3}) \ M s^{-1} = 2.5 \times 4.56 \times 10^{-3} \ M s^{-1} = 11.4 \times 10^{-3} \ M s^{-1} = 1.14 \times 10^{-2} \ M s^{-1}$.

(D) The rate expression for the reaction $x A \longrightarrow y B$ is given by: $-\frac{1}{x} \frac{d[A]}{dt} = \frac{1}{y} \frac{d[B]}{dt}$.
Given the equation: $\log_{10} \left[ -\frac{d[A]}{dt} \right] = \log_{10} \left[ \frac{d[B]}{dt} \right] + 0.3010$.
Since $\log_{10}(2) \approx 0.3010$,we can write: $\log_{10} \left[ -\frac{d[A]}{dt} \right] = \log_{10} \left[ 2 \times \frac{d[B]}{dt} \right]$.
Taking antilog on both sides: $-\frac{d[A]}{dt} = 2 \frac{d[B]}{dt}$,which implies $-\frac{1}{2} \frac{d[A]}{dt} = \frac{d[B]}{dt}$.
Comparing this with the general rate expression,we get $x = 2$ and $y = 1$.
Thus,the reaction is $2 A \longrightarrow B$. For $A = C_2H_4$ and $B = C_4H_8$,the reaction is $2 C_2H_4 \longrightarrow C_4H_8$.

(A) The given reaction is $2N_2O_{5(g)} \to 4NO_{2(g)} + O_{2(g)}$.
Rate of reaction is given by: $\text{Rate} = -\frac{1}{2} \frac{\Delta [N_2O_5]}{\Delta t} = \frac{1}{4} \frac{\Delta [NO_2]}{\Delta t}$.
Change in concentration of $N_2O_5$ is $\Delta [N_2O_5] = 2.75 - 3.00 = -0.25 \, mol \, L^{-1}$.
Time interval $\Delta t = 30 \, min$.
Rate of disappearance of $N_2O_5 = -\frac{\Delta [N_2O_5]}{\Delta t} = -\frac{-0.25}{30} = 8.333 \times 10^{-3} \, mol \, L^{-1} \, min^{-1}$.
From the stoichiometry,$\frac{1}{4} \frac{\Delta [NO_2]}{\Delta t} = -\frac{1}{2} \frac{\Delta [N_2O_5]}{\Delta t}$.
Therefore,$\frac{\Delta [NO_2]}{\Delta t} = 2 \times (-\frac{\Delta [N_2O_5]}{\Delta t}) = 2 \times 8.333 \times 10^{-3} = 1.667 \times 10^{-2} \, mol \, L^{-1} \, min^{-1}$.

(B) The rate of formation of $NH_3$ is given by: $\frac{d[NH_3]}{dt} = \frac{\Delta [NH_3]}{\Delta t} = \frac{0.04 \ M - 0.01 \ M}{20 \ s} = \frac{0.03 \ M}{20 \ s} = 1.5 \times 10^{-3} \ M \ s^{-1}$.
For the reaction $N_2 + 3H_2 \to 2NH_3$,the rate of reaction $(r)$ is defined as: $r = \frac{1}{2} \frac{d[NH_3]}{dt}$.
Substituting the value: $r = \frac{1}{2} \times 1.5 \times 10^{-3} \ M \ s^{-1} = 0.75 \times 10^{-3} \ M \ s^{-1}$.

(C) The rate of reaction for $2P + Q \to R$ is given by the expression:
$Rate = -\frac{1}{2} \frac{d[P]}{dt} = -\frac{d[Q]}{dt} = \frac{d[R]}{dt}$
Given that the rate of appearance of $R$ is $\frac{d[R]}{dt} = 0.5 \ mol \ L^{-1} \ h^{-1}$.
For $P$: $-\frac{1}{2} \frac{d[P]}{dt} = 0.5 \implies -\frac{d[P]}{dt} = 2 \times 0.5 = 1 \ mol \ L^{-1} \ h^{-1}$.
For $Q$: $-\frac{d[Q]}{dt} = 0.5 \ mol \ L^{-1} \ h^{-1}$.
Thus,the rate of disappearance of $P$ is $1 \ mol \ L^{-1} \ h^{-1}$ and for $Q$ is $0.5 \ mol \ L^{-1} \ h^{-1}$.

(D) The instantaneous rate of reaction at a specific time is determined by the slope of the tangent drawn to the concentration-time curve at that point.
For the $40^{th}$ second,the tangent passes through points corresponding to $(30, C_2)$ and $(50, C_5)$.
The slope of this tangent is calculated as $\frac{\Delta C}{\Delta t} = \frac{C_5 - C_2}{50 - 30}$.
Option $(A)$ represents this slope correctly.
Options $(B)$ and $(C)$ represent slopes of secant lines that approximate the instantaneous rate at $40 \ s$ by using points on either side of $40 \ s$ (e.g.,$(30, C_2)$ to $(40, C_3)$ or $(20, C_1)$ to $(40, C_3)$).
Option $(D)$,$\frac{C_4 - C_2}{50 - 30}$,uses $C_4$ instead of $C_5$,which does not correspond to the tangent line at $40 \ s$. Therefore,it does not represent the instantaneous rate.

(D) The rate of decomposition of $A$ is given by: $\frac{-\Delta[A]}{\Delta t} = - \frac{(4 \ M - 5 \ M)}{5 \ sec} = \frac{1 \ M}{5 \ sec} = 0.2 \ M \ sec^{-1}$.
According to the stoichiometry of the reaction,the rate of reaction is $\frac{1}{n_1} \left( \frac{-\Delta[A]}{\Delta t} \right) = \frac{1}{m_2} \left( \frac{+\Delta[D]}{\Delta t} \right)$.
Therefore,the rate of formation of $D$ is: $\frac{+\Delta[D]}{\Delta t} = \frac{m_2}{n_1} \left( \frac{-\Delta[A]}{\Delta t} \right) = \left( \frac{m_2}{n_1} \times 0.2 \right) \ M \ sec^{-1}$.

(C) The rate of reaction $r$ is given by: $r = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
Given the rate of formation of $NH_3$ is $0.001 \ kg \ L^{-1} \ h^{-1}$.
To find the rate of consumption of $H_2$ in terms of mass,we use the stoichiometry: $\frac{d[H_2]}{dt} = \frac{3}{2} \times \frac{d[NH_3]}{dt}$.
Since the rate is given in $kg \ L^{-1} \ h^{-1}$,we can directly use the stoichiometric coefficients: $\text{Rate of consumption of } H_2 = \frac{3}{2} \times (\text{Rate of formation of } NH_3) \times \frac{M_{H_2}}{M_{NH_3}}$.
$M_{H_2} = 2 \ g/mol$ and $M_{NH_3} = 17 \ g/mol$.
Rate of consumption of $H_2 = \frac{3}{2} \times 0.001 \times \frac{2}{17} \approx 0.000176 \ kg \ L^{-1} \ h^{-1}$.
Rounding to the nearest provided option,the answer is $0.00017 \ kg \ L^{-1} \ h^{-1}$.

(B) For the reaction $xA \longrightarrow yB$,the rate of reaction $(ROR)$ is given by:
$ROR = -\frac{1}{x} \frac{d[A]}{dt} = \frac{1}{y} \frac{d[B]}{dt}$
Rearranging the terms,we get:
$-\frac{d[A]}{dt} = \frac{x}{y} \frac{d[B]}{dt}$
Taking the logarithm on both sides:
$\log \left[ -\frac{d[A]}{dt} \right] = \log \left( \frac{x}{y} \cdot \frac{d[B]}{dt} \right)$
Using the property $\log(ab) = \log a + \log b$:
$\log \left[ -\frac{d[A]}{dt} \right] = \log \frac{x}{y} + \log \left[ \frac{d[B]}{dt} \right]$ ...... $(i)$
Given in the question:
$\log \left[ -\frac{d[A]}{dt} \right] = \log \left[ \frac{d[B]}{dt} \right] + \log 2$ ...... $(ii)$
Comparing equation $(i)$ and $(ii)$:
$\log \frac{x}{y} = \log 2$
Therefore,$\frac{x}{y} = 2$,which implies $x : y = 2 : 1$.

(A) The rate of reaction is given by the expression: $\text{Rate} = -\frac{1}{2} \frac{d[MnO_4^-]}{dt} = \frac{1}{5} \frac{d[I_2]}{dt}$.
Given that the rate of disappearance of $MnO_4^-$ is $-\frac{d[MnO_4^-]}{dt} = 4.56 \times 10^{-3} \ Ms^{-1}$.
Substituting this into the rate expression: $\frac{1}{2} \times (4.56 \times 10^{-3}) = \frac{1}{5} \frac{d[I_2]}{dt}$.
Therefore,the rate of appearance of $I_2$ is $\frac{d[I_2]}{dt} = \frac{5}{2} \times 4.56 \times 10^{-3} \ Ms^{-1} = 1.14 \times 10^{-2} \ Ms^{-1}$.

(B) Given reaction: $A_{2(g)} \longrightarrow B_{(g)} + \frac{1}{2} C_{(g)}$
Initial pressure at $t = 0$: $P_{A_2} = 100 \text{ mm}$,$P_B = 0$,$P_C = 0$
Pressure at $t = 5 \text{ min}$: $P_{A_2} = 100 - x$,$P_B = x$,$P_C = \frac{1}{2} x$
Total pressure $P_t = (100 - x) + x + \frac{1}{2} x = 100 + \frac{1}{2} x$
Given $P_t = 120 \text{ mm}$,so $100 + \frac{1}{2} x = 120$
$\frac{1}{2} x = 20 \implies x = 40 \text{ mm}$
The rate of disappearance of $A_2$ is defined as the change in its partial pressure over time: $\frac{x}{t} = \frac{40 \text{ mm}}{5 \text{ min}} = 8 \text{ mm min}^{-1}$.

(A) For a gaseous reaction,the concentration of $A$ is given by $[A] = \frac{n_A}{V}$.
Taking the derivative with respect to time $t$,we get $-\frac{d[A]}{dt} = -\frac{1}{V} \frac{dn_A}{dt}$.
From the ideal gas law,$P_A V = n_A RT$,so $P_A = \frac{n_A}{V} RT = [A] RT$.
Therefore,$[A] = \frac{P_A}{RT}$.
Substituting this into the rate expression,we get $-\frac{d[A]}{dt} = -\frac{d}{dt} \left( \frac{P_A}{RT} \right) = -\frac{1}{RT} \frac{dP_A}{dt}$ (assuming $T$ is constant).
Thus,the correct relation is $-\frac{d[A]}{dt} = -\frac{1}{V} \frac{dn_A}{dt} = -\frac{1}{RT} \frac{dP_A}{dt}$.

(A) For a general chemical reaction $aA + bB \to cC + dD$,the rate of reaction $(r)$ is given by the expression:
$r = - \frac{1}{a} \frac{d[A]}{dt} = - \frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$
Given the reaction $2A + 3B \to 4C$,we identify the stoichiometric coefficients as $a = 2$,$b = 3$,and $c = 4$.
Substituting these values into the rate expression,we get:
$r = - \frac{1}{2} \frac{d[A]}{dt} = - \frac{1}{3} \frac{d[B]}{dt} = \frac{1}{4} \frac{d[C]}{dt}$
Thus,option $A$ is the correct representation.

(C) The given reaction is: $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$
The rate of appearance of $NH_3$ is given as $\frac{d[NH_3]}{dt} = 2.5 \times 10^{-4} \ mol \ L^{-1} s^{-1}$.
The rate of reaction $(ROR)$ is expressed as:
$ROR = -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$
Calculating $ROR$:
$ROR = \frac{1}{2} \times (2.5 \times 10^{-4}) = 1.25 \times 10^{-4} \ mol \ L^{-1} s^{-1}$.
Calculating the rate of disappearance of $H_2$ $(-\frac{d[H_2]}{dt})$:
$-\frac{d[H_2]}{dt} = \frac{3}{2} \times \frac{d[NH_3]}{dt} = \frac{3}{2} \times (2.5 \times 10^{-4}) = 3.75 \times 10^{-4} \ mol \ L^{-1} s^{-1}$.
Thus,the rate of reaction is $1.25 \times 10^{-4} \ mol \ L^{-1} s^{-1}$ and the rate of disappearance of $H_2$ is $3.75 \times 10^{-4} \ mol \ L^{-1} s^{-1}$.
Therefore,option $C$ is correct.

(C) For any chemical reaction,the rate of reaction is defined as the change in concentration or partial pressure of a reactant or product per unit time.
For a gaseous reaction,the rate is expressed as the change in partial pressure $(P)$ per unit time $(t)$.
Therefore,the unit of the rate of reaction is $\frac{\text{pressure}}{\text{time}}$,which is $atm \cdot sec^{-1}$ or $atm \cdot s^{-1}$.
Note: While the rate constant $(k)$ for a first-order reaction has units of $s^{-1}$,the rate of the reaction itself always has units of $\text{concentration} \cdot \text{time}^{-1}$ or $\text{pressure} \cdot \text{time}^{-1}$.

(C) The rate of appearance of $B$ is determined by the first reaction path.
For the reaction $A \rightarrow 2B$,the rate of reaction is given by $\frac{1}{2} \frac{d[B]}{dt} = K_1 [A]$.
Therefore,$\frac{d[B]}{dt} = 2 K_1 [A]$.
Given $K_1 = 2 \times 10^{-3} \ s^{-1}$ and $[A] = 2 \ M$.
$\frac{d[B]}{dt} = 2 \times (2 \times 10^{-3} \ s^{-1}) \times (2 \ M) = 8 \times 10^{-3} \ M \ s^{-1}$.

(A) The rate of reaction is given by the expression: $-\frac{1}{4} \frac{d[NH_3]}{dt} = +\frac{1}{6} \frac{d[H_2O]}{dt}$.
Given that the rate of disappearance of $NH_3$ is $-\frac{d[NH_3]}{dt} = 3.6 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.
Substituting this value into the expression: $\frac{1}{4} \times (3.6 \times 10^{-3}) = \frac{1}{6} \frac{d[H_2O]}{dt}$.
Therefore,the rate of formation of $H_2O$ is $\frac{d[H_2O]}{dt} = \frac{6}{4} \times 3.6 \times 10^{-3} = 1.5 \times 3.6 \times 10^{-3} = 5.4 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.

(D) The rate of reaction for $2X \to Y$ is defined as: $\text{Rate} = -\frac{1}{2} \frac{\Delta[X]}{\Delta t}$.
Given: $\Delta[X] = [X]_f - [X]_i = 0.38 \ M - 0.50 \ M = -0.12 \ M$.
Time interval $\Delta t = 10 \ min = 10 \times 60 \ s = 600 \ s$.
Substituting the values: $\text{Rate} = -\frac{1}{2} \times \frac{-0.12 \ M}{600 \ s} = \frac{0.12}{1200} \ M s^{-1} = 1 \times 10^{-4} \ M s^{-1}$.

(B) The average rate of reaction is given by the formula:
$\text{Average Rate} = -\frac{\Delta[\text{Ester}]}{\Delta t} = -\frac{[\text{Ester}]_{t_2} - [\text{Ester}]_{t_1}}{t_2 - t_1}$
Given $t_1 = 30 \ s$ and $t_2 = 60 \ s$,the corresponding concentrations are $[\text{Ester}]_{t_1} = 0.31 \ mol \ L^{-1}$ and $[\text{Ester}]_{t_2} = 0.17 \ mol \ L^{-1}$.
$\text{Average Rate} = -\frac{0.17 - 0.31}{60 - 30} = -\frac{-0.14}{30} = \frac{0.14}{30} \ mol \ L^{-1} \ s^{-1}$
$\text{Average Rate} = 4.67 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$

(C) The molar mass of $SO_3$ is $80 \ g/mol$. Initial moles of $SO_3 = \frac{40 \ g}{80 \ g/mol} = 0.5 \ mol$.
For the reaction $2SO_3 \to 2SO_2 + O_2$,let $x$ be the moles of $O_2$ formed.
At $t = 10 \ min$,moles of $O_2 = \frac{3.2 \ g}{32 \ g/mol} = 0.1 \ mol$.
Thus,$x = 0.1 \ mol$.
The concentration of $O_2$ formed is $[O_2] = \frac{0.1 \ mol}{0.1 \ L} = 1 \ mol \ L^{-1}$.
The rate of reaction is given by $\frac{d[O_2]}{dt} = \frac{\Delta[O_2]}{\Delta t} = \frac{1 \ mol \ L^{-1}}{10 \ min} = 0.1 \ mol \ L^{-1} \ min^{-1}$.

(B) The rate of reaction for $3A \to B + 2C$ is given by the expression: $r = -\frac{1}{3} \frac{\Delta[A]}{\Delta t}$.
Given: Initial concentration $[A]_0 = 8\,mol\,L^{-1}$,Final concentration $[A]_t = 5\,mol\,L^{-1}$,Time interval $\Delta t = 10\,min$.
Calculate the change in concentration: $\Delta[A] = [A]_t - [A]_0 = 5 - 8 = -3\,mol\,L^{-1}$.
Substitute the values into the rate equation: $r = -\frac{1}{3} \times \frac{-3\,mol\,L^{-1}}{10\,min} = \frac{1}{10} = 0.1\,mol\,L^{-1}\,min^{-1}$.

(C) For the reaction $A + B \to 2C + D$,the rate of reaction is given by:
$Rate = -\frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{2} \frac{d[C]}{dt} = \frac{d[D]}{dt}$
From this expression:
$1$. Rate of disappearance of $A = -\frac{d[A]}{dt}$
$2$. Rate of disappearance of $B = -\frac{d[B]}{dt}$
$3$. Rate of appearance of $C = \frac{d[C]}{dt}$
$4$. Rate of appearance of $D = \frac{d[D]}{dt}$
Comparing these:
- Rate of disappearance of $A = \text{Rate of disappearance of } B$ (Correct)
- Rate of disappearance of $A = \text{Rate of appearance of } D$ (Correct)
- Since $-\frac{d[B]}{dt} = \frac{1}{2} \frac{d[C]}{dt}$,it implies $\text{Rate of disappearance of } B = \frac{1}{2} \times \text{Rate of appearance of } C$ (Correct)
- Therefore,the statement $\text{Rate of disappearance of } B = 2 \times \text{Rate of appearance of } C$ is incorrect.

(B) The rate of reaction is given by the expression:
Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$
To find $d[NH_3]/dt$,we equate the terms involving $H_2$ and $NH_3$:
$-\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$
Multiplying both sides by $2$:
$\frac{d[NH_3]}{dt} = -\frac{2}{3} \frac{d[H_2]}{dt}$

(C) The rate of reaction for $A_{(g)} + 3B_{(g)} \to 2C_{(g)}$ is given by the expression:
Rate $= -\frac{d[A]}{dt} = -\frac{1}{3} \frac{d[B]}{dt} = \frac{1}{2} \frac{d[C]}{dt}$
Given that $-\frac{d[A]}{dt} = 3 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$.
Equating the rates of $A$ and $B$:
$-\frac{d[A]}{dt} = -\frac{1}{3} \frac{d[B]}{dt}$
$3 \times 10^{-3} = -\frac{1}{3} \frac{d[B]}{dt}$
$-\frac{d[B]}{dt} = 3 \times (3 \times 10^{-3}) = 9 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$.

(C) The rate of reaction for $4KClO_3 \to 3KClO_4 + KCl$ is given by the expression:
Rate $= -\frac{1}{4} \frac{d[KClO_3]}{dt} = \frac{1}{3} \frac{d[KClO_4]}{dt} = \frac{d[KCl]}{dt}$.
Substituting the given rate expressions:
$-\frac{1}{4} (-K_1[KClO_3]^4) = \frac{1}{3} (K_2[KClO_3]^4) = K_3[KClO_3]^4$.
This simplifies to:
$\frac{K_1}{4} = \frac{K_2}{3} = K_3$.
Multiplying throughout by $12$,we get:
$3K_1 = 4K_2 = 12K_3$.

(A) The rate of reaction is defined as the change in concentration (or moles) over time.
Rate $= -\frac{\Delta n}{\Delta t} = -\frac{n_2 - n_1}{t_2 - t_1}$
Given: $n_1 = 2.2 \, mol$,$n_2 = 1.4 \, mol$,$\Delta t = 50 \, min = 50 \times 60 \, s = 3000 \, s$.
Rate $= -\frac{1.4 - 2.2}{3000} = \frac{0.8}{3000} \, mol \, s^{-1}$.
Rate $= 2.666... \times 10^{-4} \approx 2.67 \times 10^{-4} \, mol \, s^{-1}$.

(B) The initial rate is $r_1 = K[x][y]$.
When the volume of the vessel is reduced to $1/4$ of its initial volume $(V_2 = V_1/4)$,the concentration of the reactants increases by a factor of $4$ because concentration $C = n/V$.
Therefore,the new concentrations are $[x]_2 = 4[x]$ and $[y]_2 = 4[y]$.
The new rate $r_2 = K[x]_2[y]_2 = K(4[x])(4[y]) = 16K[x][y]$.
Thus,$r_2 = 16r_1$.
The rate of the reaction increases by $16$ times.

(B) The reaction involves the consumption of $OH^-$ ions in the first step.
Adding $HCl$ introduces $H^+$ ions into the solution.
The $H^+$ ions react with $OH^-$ ions according to the neutralization reaction: $H^+_{(aq)} + OH^-_{(aq)} \to H_2O_{(l)}$.
This decreases the concentration of $OH^-$ ions in the reaction mixture.
Since $OH^-$ is a reactant in the first step,the rate of the reaction,which depends on the concentration of $OH^-$,will decrease.

(C) In some chemical reactions,one of the products formed acts as a catalyst for the reaction.
As the reaction proceeds,the concentration of this product increases,which in turn increases the rate of the reaction.
This phenomenon is known as $Autocatalysis$.

(D) Ionic reactions involve the interaction of oppositely charged ions in solution. Because these ions exert strong electrostatic forces of attraction on each other,they combine almost immediately upon mixing. Therefore,ionic reactions are considered to be instantaneous. The Assertion is incorrect,while the Reason is correct.

(A) For a general reaction $aA + bB \to cC + dD$,the rate of reaction is given by:
Rate $= -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}$.
For the given reaction $2N_2O_5 \to 4NO_2 + O_2$,the rate of reaction is:
Rate $= -\frac{1}{2}\frac{d[N_2O_5]}{dt} = \frac{1}{4}\frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$.

(C) The rate of reaction is defined as the change in concentration of a reactant or product per unit time. Thus,the Assertion is correct.
However,the rate of reaction generally depends on the concentration of the reactants. As the reaction proceeds,the concentration of reactants decreases,which leads to a decrease in the rate of reaction. Therefore,the rate of reaction does not remain constant. Thus,the Reason is incorrect.

(C) For the reaction $N_{2(g)} + 3 H_{2(g)} \rightleftharpoons 2 NH_{3(g)}$,the rate of reaction is expressed as:
$-\frac{d[N_{2}]}{dt} = -\frac{1}{3} \frac{d[H_{2}]}{dt} = +\frac{1}{2} \frac{d[NH_{3}]}{dt}$.
Comparing this with the given options,we see that $-\frac{d[N_{2}]}{dt} = \frac{1}{2} \frac{d[NH_{3}]}{dt}$ is the correct relationship.

(N/A) The average rate of reaction is calculated using the formula: $r_{av} = -\frac{\Delta[R]}{\Delta t} = -\frac{[R]_{2} - [R]_{1}}{t_{2} - t_{1}}$.

$Time \ Interval \ (s)$	$Average \ Rate \ (mol \ L^{-1} s^{-1})$
$0-50$	$1.90 \times 10^{-4}$
$50-100$	$1.70 \times 10^{-4}$
$100-150$	$1.58 \times 10^{-4}$
$150-200$	$1.40 \times 10^{-4}$
$200-300$	$1.22 \times 10^{-4}$
$300-400$	$1.10 \times 10^{-4}$
$700-800$	$0.40 \times 10^{-4}$

(N/A) Average Rate $= \frac{1}{2} \left\{ -\frac{\Delta[N_{2}O_{5}]}{\Delta t} \right\} = -\frac{1}{2} \left[ \frac{(2.08-2.33) \ mol \ L^{-1}}{184 \ min} \right] = 6.79 \times 10^{-4} \ mol \ L^{-1} \ min^{-1}$
In terms of hours:
Rate $= (6.79 \times 10^{-4} \ mol \ L^{-1} \ min^{-1}) \times (60 \ min / 1 \ h) = 4.07 \times 10^{-2} \ mol \ L^{-1} \ h^{-1}$
In terms of seconds:
Rate $= (6.79 \times 10^{-4} \ mol \ L^{-1} \ min^{-1}) / (60 \ s / 1 \ min) = 1.13 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$
Rate of production of $NO_{2}$:
Rate $= \frac{1}{4} \frac{\Delta[NO_{2}]}{\Delta t} \implies \frac{\Delta[NO_{2}]}{\Delta t} = 4 \times \text{Average Rate}$
$= 4 \times 6.79 \times 10^{-4} \ mol \ L^{-1} \ min^{-1} = 2.72 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$

The average rate of reaction is given by the formula: $\text{Average rate} = -\frac{\Delta[R]}{\Delta t}$
Substituting the given values:
$\text{Average rate} = -\frac{[R]_2 - [R]_1}{t_2 - t_1} = -\frac{0.02 \, M - 0.03 \, M}{25 \, min} = -\frac{-0.01 \, M}{25 \, min} = 4 \times 10^{-4} \, M \, min^{-1}$
To convert the rate into $seconds^{-1}$:
$\text{Average rate} = \frac{4 \times 10^{-4} \, M}{60 \, s} = 6.67 \times 10^{-6} \, M \, s^{-1}$

(A) The rate of reaction for $2 A \rightarrow \text{Products}$ is given by: $\text{Rate} = -\frac{1}{2} \frac{\Delta [A]}{\Delta t}$
Substituting the values: $\text{Rate} = -\frac{1}{2} \frac{[A]_{2} - [A]_{1}}{t_{2} - t_{1}}$
$\text{Rate} = -\frac{1}{2} \frac{0.4 - 0.5}{10}$
$\text{Rate} = -\frac{1}{2} \frac{-0.1}{10}$
$\text{Rate} = 0.005 \ mol \ L^{-1} \ min^{-1}$
$\text{Rate} = 5 \times 10^{-3} \ M \ min^{-1}$

(N/A) The factors that affect the rate of a chemical reaction are as follows:
$(i)$ Concentration of reactants (or pressure in case of gaseous reactants).
$(ii)$ Temperature of the reaction system.
$(iii)$ Presence of a catalyst.
$(iv)$ Surface area of reactants (in case of solid reactants).
$(v)$ Exposure to radiation (in case of photochemical reactions).

(A) The average rate of reaction between the time interval $30 \ s$ to $60 \ s$ is given by the formula:
$\text{Average rate} = -\frac{\Delta [A]}{\Delta t}$
$= -\frac{[A]_{60} - [A]_{30}}{60 - 30}$
$= -\frac{0.17 - 0.31}{30}$
$= -\frac{-0.14}{30}$
$= 4.67 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$