Rate of a reaction Questions in English

(B) The rate of a chemical reaction is proportional to the concentration of the reactants.
As the reaction proceeds,the concentration of the reactants decreases over time.
Therefore,the rate of reaction continuously decreases with time.

(A) The rate of a reaction depends on factors such as the concentration of reactants,surface area,temperature,presence of light,and catalysts. $Pressure$ only affects the rate of reactions involving gaseous reactants or products because it changes the concentration of gases. Therefore,for a reaction that does not involve gases,the rate is not dependent on $Pressure$.

(C) The rate of reaction is defined as the change in concentration of the reactant over time.
Rate $= -\frac{\Delta[R]}{\Delta t} = -\frac{[R]_f - [R]_i}{t_f - t_i}$
Given: $[R]_i = 0.2 \ M$,$[R]_f = 0.1 \ M$,$\Delta t = 10 \ min$
Rate $= -\frac{0.1 - 0.2}{10} = \frac{0.1}{10} = 0.01 \ mol \ dm^{-3} \ min^{-1}$
Therefore,the correct option is $C$.

(A) The balanced chemical equation for Haber's process is: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$.
The rate of reaction is expressed as: $-\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt}$.
Given that the rate of formation of ammonia is $\frac{d[NH_3]}{dt} = 40 \times 10^{-3} \ mol \ L^{-1} s^{-1}$.
To find the rate of disappearance of hydrogen $(-\frac{d[H_2]}{dt})$,we use the relation: $-\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt}$.
$-\frac{d[H_2]}{dt} = \frac{3}{2} \times \frac{d[NH_3]}{dt} = \frac{3}{2} \times 40 \times 10^{-3} \ mol \ L^{-1} s^{-1} = 60 \times 10^{-3} \ mol \ L^{-1} s^{-1}$.

(B) Ionic reactions involve the interaction of ions in solution,which occurs almost instantaneously.
Therefore,the time required for the completion of ionic reactions is much less compared to molecular reactions,which involve the breaking and formation of covalent bonds.

(C) For the reaction $2A + B \to A_2B$,the rate of reaction is given by:
$Rate = -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{d[A_2B]}{dt}$
From the expression $-\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt}$,we get:
$-\frac{d[A]}{dt} = 2 \times (-\frac{d[B]}{dt})$
This shows that the rate of disappearance of $A$ $(-\frac{d[A]}{dt})$ is twice the rate of disappearance of $B$ $(-\frac{d[B]}{dt})$.

(B) The rate of reaction is given by the expression: $\text{Rate} = -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
Given that the rate of formation of $NH_3$ is $\frac{d[NH_3]}{dt} = 0.001 \ kg \ h^{-1}$.
From the stoichiometric relationship: $-\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
Therefore,the rate of disappearance (conversion) of $H_2$ is: $-\frac{d[H_2]}{dt} = \frac{3}{2} \times \frac{d[NH_3]}{dt} = 1.5 \times 0.001 \ kg \ h^{-1} = 0.0015 \ kg \ h^{-1}$.

(B) In a chemical reaction,the rate of reaction with respect to the reactant is given by the negative change in concentration over time.
Therefore,the term $\left( -\frac{dc}{dt} \right)$ represents the decrease in the concentration of the reactant with respect to time.

(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 reaction $3A + B \to C + D$,the stoichiometric coefficients are $a=3$,$b=1$,$c=1$,and $d=1$.
Substituting these values,we get:
Rate $= -\frac{1}{3}\frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{d[C]}{dt} = \frac{d[D]}{dt}$.
Thus,option $A$ is correct.

(B) The rate of reaction for $N_2 + 3H_2 \to 2NH_3$ is given by:
$-\frac{\Delta [N_2]}{\Delta t} = -\frac{1}{3} \frac{\Delta [H_2]}{\Delta t} = \frac{1}{2} \frac{\Delta [NH_3]}{\Delta t}$
From the relation $-\frac{1}{3} \frac{\Delta [H_2]}{\Delta t} = \frac{1}{2} \frac{\Delta [NH_3]}{\Delta t}$,we get:
$-\frac{\Delta [H_2]}{\Delta t} = \frac{3}{2} \times \frac{\Delta [NH_3]}{\Delta t}$
Substituting the given value $\frac{\Delta [NH_3]}{\Delta t} = 2 \times 10^{-4} \ mol \ L^{-1} s^{-1}$:
$-\frac{\Delta [H_2]}{\Delta t} = \frac{3}{2} \times 2 \times 10^{-4} \ mol \ L^{-1} s^{-1} = 3 \times 10^{-4} \ mol \ L^{-1} s^{-1}$

(A) The rate of appearance of a product is defined as the change in its concentration per unit time.
Given: Increase in concentration of $B = 5 \times 10^{-3} \ mol \ L^{-1}$.
Time taken = $10 \ s$.
Rate of appearance of $B = \frac{\Delta [B]}{\Delta t} = \frac{5 \times 10^{-3} \ mol \ L^{-1}}{10 \ s} = 5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.

(D) The balanced chemical equation is $2SO_2 + O_2 \to 2SO_3$.
According to the stoichiometry of the reaction,$2 \text{ moles}$ of $SO_2$ produce $2 \text{ moles}$ of $SO_3$.
Since the molar mass of $SO_2$ $(64 \ g/mol)$ is equal to the molar mass of $SO_3$ $(80 \ g/mol)$ is not the case,we must relate the mass rates.
Rate of reaction $= -\frac{1}{2} \frac{d[SO_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt}$.
In terms of mass rate: $-\frac{1}{2 \times M_{SO_2}} \frac{d(mass_{SO_2})}{dt} = \frac{1}{2 \times M_{SO_3}} \frac{d(mass_{SO_3})}{dt}$.
Given $-\frac{d(mass_{SO_2})}{dt} = 1.28 \times 10^{-3} \ g/sec$.
$\frac{d(mass_{SO_3})}{dt} = \frac{M_{SO_3}}{M_{SO_2}} \times (\text{Rate of disappearance of } SO_2) = \frac{80}{64} \times 1.28 \times 10^{-3} \ g/sec$.
$\frac{d(mass_{SO_3})}{dt} = 1.25 \times 1.28 \times 10^{-3} \ g/sec = 1.60 \times 10^{-3} \ g/sec$.

(A) The given reaction is $2N_2O_{5(g)} \to 4NO_{2(g)} + O_{2(g)}$.
The rate of reaction is expressed in terms of the rate of change of concentration of products as:
$\text{Rate} = \frac{1}{4} \frac{d[NO_2]}{dt}$
Given that $\frac{d[NO_2]}{dt} = \frac{5.2 \times 10^{-3} \ M}{100 \ s} = 5.2 \times 10^{-5} \ M \ s^{-1}$.
Substituting this value into the rate expression:
$\text{Rate} = \frac{1}{4} \times (5.2 \times 10^{-5} \ M \ s^{-1}) = 1.3 \times 10^{-5} \ M \ s^{-1}$.

(A) The rate of reaction for $A + 2B \to C + D$ is given by the expression:
Rate $= -\frac{d[A]}{dt} = -\frac{1}{2} \frac{d[B]}{dt} = \frac{d[C]}{dt} = \frac{d[D]}{dt}$
Given that $-\frac{d[A]}{dt} = 5 \times 10^{-4} \ mol \ L^{-1} s^{-1}$,we can equate this to the rate of disappearance of $B$:
$-\frac{d[A]}{dt} = -\frac{1}{2} \frac{d[B]}{dt}$
$-\frac{d[B]}{dt} = 2 \times (-\frac{d[A]}{dt})$
$-\frac{d[B]}{dt} = 2 \times (5 \times 10^{-4} \ mol \ L^{-1} s^{-1}) = 1.0 \times 10^{-3} \ mol \ L^{-1} s^{-1}$

(A) For a general reaction $aA + bB \rightarrow cC + dD$,the rate of reaction is given by:
Rate $= -\frac{1}{a} \frac{\Delta [A]}{\Delta t} = -\frac{1}{b} \frac{\Delta [B]}{\Delta t} = \frac{1}{c} \frac{\Delta [C]}{\Delta t} = \frac{1}{d} \frac{\Delta [D]}{\Delta t}$.
For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,the stoichiometric coefficients are $1, 1, 2$ respectively.
Thus,the rate of reaction is expressed as:
Rate $= -\frac{\Delta [H_2]}{\Delta t} = -\frac{\Delta [I_2]}{\Delta t} = \frac{1}{2} \frac{\Delta [HI]}{\Delta t}$.

(C) For the reaction $3A \to 2B$,the rate of reaction is expressed as:
Rate = $-\frac{1}{3}\frac{d[A]}{dt} = +\frac{1}{2}\frac{d[B]}{dt}$
To find the rate of formation of $B$,which is $+\frac{d[B]}{dt}$,we multiply both sides by $2$:
$+\frac{d[B]}{dt} = -\frac{2}{3}\frac{d[A]}{dt}$

(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 reaction $H_2 + I_2 \to 2HI$,the stoichiometric coefficients are $1$ for $H_2$,$1$ for $I_2$,and $2$ for $HI$.
Therefore,the rate of reaction is:
$Rate = -\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = +\frac{1}{2}\frac{d[HI]}{dt}$.

(D) The correct option is $(D)$.
Combustion is an exothermic process,but the rate of reaction generally increases with an increase in temperature due to the increase in the number of effective collisions.
For the combustion of carbon,higher temperatures facilitate the burning of more particles per unit time.
Therefore,the reaction at $1000 \ ^oC$ will be the fastest among the given options.

(B) The reaction is an acid-catalyzed hydrolysis of an ester.
As the reaction progresses,the concentration of $CH_3COOH$ (acetic acid) increases.
This increase in the concentration of acetic acid can be determined by titrating the reaction mixture against a standard alkali solution at different time intervals.
Therefore,the progress of the reaction is followed by finding the amount of acetic acid formed at different intervals.

(B) 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 ${N_2}_{(g)} + 3{H_2}_{(g)} \to 2N{H_3}_{(g)}$,the rate expression is:
Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[{H_2}]}{dt} = \frac{1}{2}\frac{d[N{H_3}]}{dt}$.
Equating the terms for $N{H_3}$ and ${H_2}$:
$\frac{1}{2}\frac{d[N{H_3}]}{dt} = -\frac{1}{3}\frac{d[{H_2}]}{dt}$.
Multiplying both sides by $2$:
$\frac{d[N{H_3}]}{dt} = -\frac{2}{3}\frac{d[{H_2}]}{dt}$.

(B) The rate of reaction is given by: $Rate = -\frac{d[N_2O_4]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt}$
Given,change in pressure of $N_2O_4$ is $\Delta P = 0.28 - 0.46 = -0.18 \ atm$ over $\Delta t = 30 \ min$.
The rate of disappearance of $N_2O_4$ is $-\frac{\Delta P}{\Delta t} = -\frac{-0.18}{30} = 0.006 \ atm \ min^{-1}$.
Equating the rates: $\frac{1}{2} \frac{d[NO_2]}{dt} = 0.006 \ atm \ min^{-1}$.
Therefore,the rate of formation of $NO_2$ is $\frac{d[NO_2]}{dt} = 2 \times 0.006 = 0.012 \ atm \ min^{-1} = 1.2 \times 10^{-2} \ atm \ min^{-1}$.

(B) For the reaction $4A + B \rightarrow 2C + 2D$,the rate of reaction is given by:
$-\frac{1}{4} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{2} \frac{d[C]}{dt} = \frac{1}{2} \frac{d[D]}{dt}$.
From this,we can derive:
$1$. Rate of disappearance of $B$ $(-\frac{d[B]}{dt})$ is $1/4$ of the rate of disappearance of $A$ $(-\frac{d[A]}{dt})$. This is correct.
$2$. Rate of appearance of $C$ $(\frac{d[C]}{dt})$ is $2 \times (-\frac{d[B]}{dt})$. The statement says it is half,which is incorrect.
$3$. Rate of formation of $D$ $(\frac{d[D]}{dt})$ is $\frac{1}{2} \times (-\frac{d[A]}{dt})$. This is correct.
$4$. Rate of formation of $C$ and $D$ are equal $(\frac{d[C]}{dt} = \frac{d[D]}{dt})$. This is correct.
Therefore,the incorrect statement is $B$.

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

(A) The rate of reaction $(r)$ is defined as the rate of disappearance of reactants divided by their stoichiometric coefficients.
For the reaction $2A + 3B \rightarrow \text{Products}$,the rate is given by:
$r = -\frac{1}{2} \frac{d[A]}{dt} = -\frac{1}{3} \frac{d[B]}{dt}$
Given that the rate of disappearance of $A$ is $r_1 = -\frac{d[A]}{dt}$ and the rate of disappearance of $B$ is $r_2 = -\frac{d[B]}{dt}$,we substitute these into the expression:
$r = \frac{1}{2} r_1 = \frac{1}{3} r_2$
Multiplying by $6$,we get:
$3r_1 = 2r_2$

(C) The rate of reaction is defined as the rate of change of concentration of any reactant or product divided by its stoichiometric coefficient.
For the given reaction: $6H^+ + BrO_3^- + 5Br^- \to 3Br_2 + 3H_2O$
The rate of reaction is given by:
Rate $= -\frac{1}{6} \frac{d[H^+]}{dt} = -\frac{d[BrO_3^-]}{dt} = -\frac{1}{5} \frac{d[Br^-]}{dt} = \frac{1}{3} \frac{d[Br_2]}{dt} = \frac{1}{3} \frac{d[H_2O]}{dt}$
Comparing the terms for $Br^-$ and $Br_2$:
$-\frac{1}{5} \frac{d[Br^-]}{dt} = \frac{1}{3} \frac{d[Br_2]}{dt}$
Multiplying both sides by $5$,we get:
$-\frac{d[Br^-]}{dt} = \frac{5}{3} \frac{d[Br_2]}{dt}$

(D) The balanced chemical equation is $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$.
According to the rate law,the rate of reaction is given by: $Rate = -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
Given the rate of formation of $NH_3$ is $0.001 \, kg \, h^{-1}$,we convert this to molar rate: $Molar \, mass \, of \, NH_3 = 17 \, g/mol$ and $H_2 = 2 \, g/mol$.
Rate of formation of $NH_3$ in $mol \, h^{-1} = \frac{0.001 \times 1000}{17} = \frac{1}{17} \, mol \, h^{-1}$.
From the stoichiometry,$-\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
So,$-\frac{d[H_2]}{dt} = \frac{3}{2} \times \frac{d[NH_3]}{dt} = \frac{3}{2} \times \frac{1}{17} = \frac{3}{34} \, mol \, h^{-1}$.
Converting back to $kg \, h^{-1}$: $Rate \, of \, consumption \, of \, H_2 = \frac{3}{34} \times 2 \, g \, h^{-1} = \frac{6}{34} \, g \, h^{-1} \approx 0.176 \, g \, h^{-1} = 0.000176 \, kg \, h^{-1}$.
Rounding to the nearest option,the value is $0.0017 \, kg \, h^{-1}$.

(C) For a general reaction $aA + bB \rightarrow cC + dD$,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 $H_2 + I_2 \rightleftharpoons 2HI$,the stoichiometric coefficients are $1, 1,$ and $2$ respectively.
Therefore,the rate of reaction is:
Rate $= -\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = \frac{1}{2} \frac{d[HI]}{dt}$.
Rearranging this,we get $-\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = \frac{1}{2} \frac{d[HI]}{dt}$.

(B) For a general reaction $aA \to bB$,the rate of reaction is given by:
Rate $= -\frac{1}{a} \frac{d[A]}{dt} = +\frac{1}{b} \frac{d[B]}{dt}$
Given the reaction $3A \to 2B$,we have $a=3$ and $b=2$.
Substituting these values:
$-\frac{1}{3} \frac{d[A]}{dt} = +\frac{1}{2} \frac{d[B]}{dt}$
To find $\frac{d[B]}{dt}$,multiply both sides by $2$:
$\frac{d[B]}{dt} = -\frac{2}{3} \frac{d[A]}{dt}$

(B) The rate of reaction is given by the expression: $\text{Rate} = \frac{1}{4} \frac{d[NO_2]}{dt}$.
Given that the change in concentration of $NO_2$ is $\Delta[NO_2] = 2.0 \times 10^{-2} \, mol \, L^{-1}$ and the time interval is $\Delta t = 5 \, s$.
Therefore,$\frac{d[NO_2]}{dt} = \frac{2.0 \times 10^{-2} \, mol \, L^{-1}}{5 \, s} = 4 \times 10^{-3} \, mol \, L^{-1} \, s^{-1}$.
Substituting this into the rate expression: $\text{Rate} = \frac{1}{4} \times (4 \times 10^{-3} \, mol \, L^{-1} \, s^{-1}) = 10^{-3} \, mol \, L^{-1} \, s^{-1}$.

(A) For a general reaction $aA + bB \rightarrow cC + dD$,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 $2A + B \rightarrow 3C + D$,the rate is:
Rate $= -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{3} \frac{d[C]}{dt} = \frac{d[D]}{dt}$.
Comparing this with the options,option $A$ is $-\frac{1}{3} \frac{d[C]}{dt}$,which is incorrect because the rate should be $+\frac{1}{3} \frac{d[C]}{dt}$ for the product $C$.

(A) For the reaction $2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)}$,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 $O_2$:
$-\frac{1}{2} \frac{d[SO_2]}{dt} = -\frac{d[O_2]}{dt}$
$-\frac{d[SO_2]}{dt} = 2 \times (-\frac{d[O_2]}{dt})$
$-\frac{d[SO_2]}{dt} = 2 \times (2.5 \times 10^{-4} \, mol \, L^{-1} \, s^{-1}) = 5 \times 10^{-4} \, mol \, L^{-1} \, s^{-1}$.

(A) The rate of formation of $NO_2$ is given by the change in concentration of $NO_2$ divided by the time interval.
Rate of formation $= \frac{\Delta[NO_2]}{\Delta t} = \frac{1.6 \times 10^{-2} \ M}{4 \ s} = 0.4 \times 10^{-2} \ M \ s^{-1} = 4 \times 10^{-3} \ M \ s^{-1}$.

(B) For a general reaction $aA + bB \rightarrow cC + dD$,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 $2A + B \rightarrow 3C + D$,the rate is:
Rate $= -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{3} \frac{d[C]}{dt} = \frac{d[D]}{dt}$.
Comparing this with the options,option $B$ is incorrect because it includes a negative sign for a product $(C)$,which is physically incorrect as the concentration of products increases over time.

(C) The rate of reaction is given by the expression:
$-\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$
Rearranging to solve for $-\frac{d[H_2]}{dt}$:
$-\frac{d[H_2]}{dt} = \frac{3}{2} \frac{d[NH_3]}{dt}$
Substituting the given value $\frac{d[NH_3]}{dt} = 2 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$:
$-\frac{d[H_2]}{dt} = \frac{3}{2} \times (2 \times 10^{-4}) = 3 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$

(C) The rate of reaction is given by: $Rate = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$.
Given,the rate of formation of $NO_2$ is $\frac{d[NO_2]}{dt} = \frac{2.0 \times 10^{-2} \ mol \ L^{-1}}{5 \ s} = 4.0 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.
From the rate expression: $-\frac{d[N_2O_5]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt}$.
Substituting the value: $-\frac{d[N_2O_5]}{dt} = \frac{1}{2} \times (4.0 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}) = 2.0 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.

(A) For a general reaction $aA + bB \rightarrow cC + dD$,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 $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,the stoichiometric coefficients are $a=1$,$b=3$,and $c=2$.
Substituting these values,we get:
Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt}$.

(B) The rate of reaction is given by the expression:
$Rate = -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt}$
Given that the rate of formation of $NH_3$ is $\frac{d[NH_3]}{dt} = 2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
Equating the rate of disappearance of $N_2$ to the rate of formation of $NH_3$:
$-\frac{d[N_2]}{dt} = \frac{1}{2} \times \frac{d[NH_3]}{dt}$
$-\frac{d[N_2]}{dt} = \frac{1}{2} \times (2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1})$
$-\frac{d[N_2]}{dt} = 1.25 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$

(B) For a general reaction $aA \rightarrow bB$,the rate of reaction is given by:
Rate $= -\frac{1}{a} \frac{d[A]}{dt} = \frac{1}{b} \frac{d[B]}{dt}$.
Given the reaction $\frac{1}{2} A \rightarrow 2B$,here $a = \frac{1}{2}$ and $b = 2$.
Substituting these values:
Rate $= -\frac{1}{1/2} \frac{d[A]}{dt} = \frac{1}{2} \frac{d[B]}{dt}$.
$-2 \frac{d[A]}{dt} = \frac{1}{2} \frac{d[B]}{dt}$.
Multiplying both sides by $\frac{1}{2}$:
$-\frac{d[A]}{dt} = \frac{1}{4} \frac{d[B]}{dt}$.

(A) For a general reaction $aA + bB \rightarrow cC + dD$,the rate of reaction is given by:
Rate $= -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}$
Applying this to the reaction $N_2 + 3H_2 \rightarrow 2NH_3$:
Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt}$
Thus,option $A$ is correct.

(D) For a general reaction $aA + bB \rightarrow cC + dD$,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 $H_2 + I_2 \rightarrow 2HI$,the stoichiometric coefficients are $1, 1,$ and $2$ respectively.
Therefore,the rate of reaction is:
$Rate = -\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = \frac{1}{2} \frac{d[HI]}{dt}$
Multiplying by $2$,we get:
$2 \left( -\frac{d[H_2]}{dt} \right) = 2 \left( -\frac{d[I_2]}{dt} \right) = \frac{d[HI]}{dt}$
Alternatively,the expression can be written as:
$-\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = \frac{1}{2} \frac{d[HI]}{dt}$
Comparing this with the given options,option $D$ represents the correct relationship.

(A) The rate of reaction is given by:
Rate $= -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$.
The rate of decomposition of $N_2O_5$ is defined as $-\frac{d[N_2O_5]}{dt}$.
The rate of formation of $NO_2$ is defined as $\frac{d[NO_2]}{dt}$.
From the rate expression: $-\frac{d[N_2O_5]}{dt} = \frac{2}{4} \frac{d[NO_2]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt}$.
Therefore,the ratio of the rate of decomposition of $N_2O_5$ to the rate of formation of $NO_2$ is $\frac{-\frac{d[N_2O_5]}{dt}}{\frac{d[NO_2]}{dt}} = \frac{1}{2}$,which is $1:2$.

(B) The rate of reaction is given by the expression: $-\frac{d[N_{2}O_{5}]}{dt} = \frac{1}{2} \frac{d[NO_{2}]}{dt} = 2 \frac{d[O_{2}]}{dt}$.
Given,$-\frac{d[N_{2}O_{5}]}{dt} = 6.25 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.
For $NO_{2}$ formation: $\frac{d[NO_{2}]}{dt} = 2 \times (-\frac{d[N_{2}O_{5}]}{dt}) = 2 \times 6.25 \times 10^{-3} = 1.25 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$.
For $O_{2}$ formation: $\frac{d[O_{2}]}{dt} = \frac{1}{2} \times (-\frac{d[N_{2}O_{5}]}{dt}) = \frac{6.25 \times 10^{-3}}{2} = 3.125 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.

(D) For the reaction,$N_2 + 3H_2 \rightarrow 2NH_3,$ the rate of reaction is given by:
Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$
Given that $\frac{d[NH_3]}{dt} = 2 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}.$
Equating the terms for $H_2$ and $NH_3$:
$-\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$
$-\frac{d[H_2]}{dt} = \frac{3}{2} \times \frac{d[NH_3]}{dt}$
$-\frac{d[H_2]}{dt} = \frac{3}{2} \times (2 \times 10^{-4}) = 3 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}.$

(D) For the reaction $BrO_{3(aq)}^{-} + 5Br_{(aq)}^{-} + 6H^{+} \to 3Br_{2(l)} + 3H_2O_{(l)}$,the rate of reaction is given by:
Rate $= -\frac{d[BrO_3^-]}{dt} = -\frac{1}{5} \frac{d[Br^-]}{dt} = -\frac{1}{6} \frac{d[H^+]}{dt} = \frac{1}{3} \frac{d[Br_2]}{dt} = \frac{1}{3} \frac{d[H_2O]}{dt}$.
We need to relate the rate of appearance of $Br_2$ to the rate of disappearance of $Br^-$.
From the expression: $\frac{1}{3} \frac{d[Br_2]}{dt} = -\frac{1}{5} \frac{d[Br^-]}{dt}$.
Multiplying both sides by $3$,we get:
$\frac{d[Br_2]}{dt} = -\frac{3}{5} \frac{d[Br^-]}{dt}$.

(C) For the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,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 $H_2$:
$+ \frac{1}{2} \frac{d[NH_3]}{dt} = - \frac{1}{3} \frac{d[H_2]}{dt}$.
Multiplying both sides by $2$,we get:
$+ \frac{d[NH_3]}{dt} = - \frac{2}{3} \frac{d[H_2]}{dt}$.

(B) For the reaction $2A + B \rightarrow 3C + D,$ the rate of reaction is expressed as:
Rate $= -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{3} \frac{d[C]}{dt} = \frac{d[D]}{dt}$
Comparing this with the given options:
Option $A$ is $-\frac{1}{2} \frac{d[A]}{dt}$ (Correct expression).
Option $B$ is $-\frac{d[C]}{3dt}$,but the correct expression for $C$ is $+\frac{1}{3} \frac{d[C]}{dt}$. Thus,this is incorrect.
Option $C$ is $-\frac{d[B]}{dt}$ (Correct expression).
Option $D$ is $\frac{d[D]}{dt}$ (Correct expression).
Therefore,the expression that does not represent the reaction rate is $-\frac{d[C]}{3dt}$.

(B) The given reaction is $2SO_2 + O_2 \to 2SO_3$.
The rate expression for this reaction is given by:
$-\frac{1}{2} \frac{d[SO_2]}{dt} = -\frac{d[O_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt}$.
We are given the rate of disappearance of $O_2$,which is $-\frac{d[O_2]}{dt} = 2.0 \times 10^{-4} \ mol \ L^{-1}s^{-1}$.
From the rate expression,we have:
$-\frac{d[O_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt}$.
Therefore,the rate of appearance of $SO_3$ is:
$\frac{d[SO_3]}{dt} = 2 \times (-\frac{d[O_2]}{dt}) = 2 \times (2.0 \times 10^{-4} \ mol \ L^{-1}s^{-1}) = 4.0 \times 10^{-4} \ mol \ L^{-1}s^{-1}$.

(C) For a general reaction $aA + bB \rightarrow cC + dD$,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 $I_2 + 2S_2O_3^{2-} \rightarrow 2I^{-} + S_4O_6^{2-}$,the rate expression is:
Rate $= -\frac{d[I_2]}{dt} = -\frac{1}{2}\frac{d[S_2O_3^{2-}]}{dt} = \frac{1}{2}\frac{d[I^{-}]}{dt} = \frac{d[S_4O_6^{2-}]}{dt}$.
Comparing this with the given options,the expression $-\frac{d[I_2]}{dt} = \frac{d[S_4O_6^{2-}]}{dt}$ is correct.

(A) The molar mass of $O_2$ is $32 \ g \ mol^{-1}$.
Rate of formation of $O_2$ in $mol \ min^{-1} = \frac{36 \ g \ min^{-1}}{32 \ g \ mol^{-1}} = 1.125 \ mol \ min^{-1}$.
From the stoichiometry of the reaction $2H_2O_2 \to 2H_2O + O_2$,the rate expression is given by $-\frac{1}{2} \frac{d[H_2O_2]}{dt} = \frac{d[O_2]}{dt}$.
Therefore,the rate of disappearance of $H_2O_2$ is $-\frac{d[H_2O_2]}{dt} = 2 \times \frac{d[O_2]}{dt}$.
Substituting the value,$-\frac{d[H_2O_2]}{dt} = 2 \times 1.125 \ mol \ min^{-1} = 2.25 \ mol \ min^{-1}$.