Rate of a reaction Questions in English

(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 reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the rate expression is:
$Rate = -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$
Thus,option $A$ is the correct representation.

(B) For the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,the rate of reaction is expressed as:
Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
From this,we can equate the rate of disappearance of $N_2$ and the rate of appearance of $NH_3$:
$-\frac{d[N_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
Thus,option $B$ is correct.

(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 $2 NO_{(g)} + 2 H_{2(g)} \rightarrow N_{2(g)} + 2 H_2O_{(g)}$,the rate expression is:
Rate $= -\frac{1}{2} \frac{d[NO]}{dt} = -\frac{1}{2} \frac{d[H_2]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{2} \frac{d[H_2O]}{dt}$.
Comparing the terms for $NO$ and $H_2O$:
$-\frac{1}{2} \frac{d[NO]}{dt} = \frac{1}{2} \frac{d[H_2O]}{dt}$.
Multiplying both sides by $2$,we get:
$-\frac{d[NO]}{dt} = \frac{d[H_2O]}{dt}$.

(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 given reaction $2 NO_{(g)} + 2 H_{2(g)} \rightarrow N_{2(g)} + 2 H_2 O_{(g)}$,we get:
$Rate = -\frac{1}{2} \frac{d[NO]}{dt} = -\frac{1}{2} \frac{d[H_2]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{2} \frac{d[H_2 O]}{dt}$.
Comparing this with the given options,option $A$ represents the correct stoichiometric relationship.

(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 $2 NO_{(g)} + 2 H_{2(g)} \rightarrow N_{2(g)} + 2 H_{2}O_{(g)}$,the rate expression is:
Rate $= -\frac{1}{2} \frac{d[NO]}{dt} = -\frac{1}{2} \frac{d[H_2]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{2} \frac{d[H_2O]}{dt}$
Comparing the terms for $N_2$ and $H_2O$:
$\frac{d[N_2]}{dt} = \frac{1}{2} \frac{d[H_2O]}{dt}$
This implies $\frac{d[H_2O]}{dt} = 2 \frac{d[N_2]}{dt}$.
Comparing the terms for $N_2$ and $H_2$:
$\frac{d[N_2]}{dt} = -\frac{1}{2} \frac{d[H_2]}{dt}$.
Thus,option $B$ is the correct relationship.

(A) The rate of reaction is given by the expression: $\text{Rate} = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$.
Given that the concentration of $NO_2$ increases by $\Delta[NO_2] = 5.2 \times 10^{-3} \ M$ in $\Delta t = 100 \ s$.
The rate of formation of $NO_2$ is $\frac{d[NO_2]}{dt} = \frac{5.2 \times 10^{-3} \ M}{100 \ s} = 5.2 \times 10^{-5} \ M \ s^{-1}$.
Therefore,the rate of the reaction is $\text{Rate} = \frac{1}{4} \times (5.2 \times 10^{-5} \ M \ s^{-1}) = 1.3 \times 10^{-5} \ mol \ dm^{-3} \ s^{-1}$.

(B) For a general reaction $aA + bB \rightarrow cC + dD$,the rate of reaction is given by: $\text{Rate} = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$.
Comparing this with the given expression $-\frac{1}{2} \frac{d[x]}{dt} = -\frac{d[y]}{dt} = \frac{1}{2} \frac{d[z]}{dt}$,we can identify the stoichiometric coefficients:
For reactant $x$,the coefficient is $2$.
For reactant $y$,the coefficient is $1$.
For product $z$,the coefficient is $2$.
Thus,the balanced chemical equation is $2x + y \rightarrow 2z$.

(A) The average rate of reaction is defined as the change in concentration of the product divided by the time interval.
Formula: $\text{Average Rate} = \frac{\Delta [\text{Product}]}{\Delta t}$
Given: $\Delta [\text{Product}] = 0.05 \ M$ and $\Delta t = 20 \ s$.
Calculation: $\text{Average Rate} = \frac{0.05 \ M}{20 \ s} = 0.0025 \ M \ s^{-1}$.
Therefore,the correct option is $A$.

(B) For the reaction $A + 3 B \rightarrow 2 C$,the rate of reaction is given by:
$-\frac{d[A]}{dt} = -\frac{1}{3} \frac{d[B]}{dt} = \frac{1}{2} \frac{d[C]}{dt}$
Given that the rate of consumption of $A$ is $-\frac{d[A]}{dt} = 1.4 \ mol \ dm^{-3} \ minute^{-1}$.
We need to find the rate of consumption of $B$,which is $-\frac{d[B]}{dt}$.
From the rate expression: $-\frac{d[A]}{dt} = -\frac{1}{3} \frac{d[B]}{dt}$
Therefore,$-\frac{d[B]}{dt} = 3 \times (-\frac{d[A]}{dt}) = 3 \times 1.4 \ mol \ dm^{-3} \ minute^{-1} = 4.2 \ mol \ dm^{-3} \ minute^{-1}$.

(C) The rate of reaction is given by the expression: $Rate = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{d[O_2]}{dt}$.
Given that the rate of disappearance of $N_2O_5$ is $-\frac{d[N_2O_5]}{dt} = x \ mol \ dm^{-3} \ s^{-1}$.
Substituting this into the rate expression: $Rate = \frac{1}{2} \times x = \frac{d[O_2]}{dt}$.
Therefore,the rate of formation of $O_2$ is $\frac{x}{2} \ mol \ dm^{-3} \ s^{-1}$.

(A) The rate of reaction is given by the expression:
$\text{Rate} = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{d[O_2]}{dt}$
Given that the rate of disappearance of $N_2O_5$ is $\left| \frac{d[N_2O_5]}{dt} \right| = 0.02 \ mol \ dm^{-3} \ s^{-1}$.
Substituting this value into the expression:
$\frac{d[O_2]}{dt} = \frac{1}{2} \times 0.02 \ mol \ dm^{-3} \ s^{-1} = 0.01 \ mol \ dm^{-3} \ s^{-1}$.

(C) For the reaction,$A + 3B \longrightarrow 2C$,the rate of reaction is given by:
$-\frac{d[A]}{dt} = -\frac{1}{3} \frac{d[B]}{dt} = \frac{1}{2} \frac{d[C]}{dt}$
Given that the rate of consumption of $A$ is $-\frac{d[A]}{dt} = 1.4 \ mol \ dm^{-3} \ s^{-1}$.
Equating the terms for $A$ and $C$:
$-\frac{d[A]}{dt} = \frac{1}{2} \frac{d[C]}{dt}$
$1.4 = \frac{1}{2} \frac{d[C]}{dt}$
$\frac{d[C]}{dt} = 2 \times 1.4 \ mol \ dm^{-3} \ s^{-1} = 2.8 \ mol \ dm^{-3} \ s^{-1}$

(B) The given reaction is: $2 \ N_2O_{5(g)} \longrightarrow 4 \ NO_{2(g)} + O_{2(g)}$.
The rate of reaction is expressed as: $-\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt}$.
Given that the rate of disappearance of $N_2O_5$ is $-\frac{d[N_2O_5]}{dt} = 0.06 \ mol \ dm^{-3} \ s^{-1}$.
Substituting this into the rate expression: $\frac{1}{2} \times (0.06) = \frac{1}{4} \frac{d[NO_2]}{dt}$.
Therefore,the rate of formation of $NO_2$ is: $\frac{d[NO_2]}{dt} = 2 \times 0.06 = 0.12 \ mol \ dm^{-3} \ s^{-1}$.

(C) The given reaction is $3 I_{(aq)}^{-} + S_2 O_{8_{(aq)}}^{2-} \longrightarrow 2 SO_{4_{(aq)}}^{2-} + I_{3_{(aq)}}^{-}$.
The rate of reaction is expressed as: $\text{Rate} = -\frac{1}{3} \frac{d[I^{-}]}{dt} = -\frac{d[S_2 O_8^{2-}]}{dt} = \frac{1}{2} \frac{d[SO_4^{2-}]}{dt} = \frac{d[I_3^{-}]}{dt}$.
From the expression,the rate of consumption of $I^{-}$ is given by $-\frac{d[I^{-}]}{dt} = \frac{3}{2} \times \frac{d[SO_4^{2-}]}{dt}$.
Given $\frac{d[SO_4^{2-}]}{dt} = 0.044 \ mol \ dm^{-3} \ s^{-1}$,we have:
Rate of consumption of $I^{-} = \frac{3}{2} \times 0.044 = 0.066 \ mol \ dm^{-3} \ s^{-1}$.

(D) The balanced chemical equation is: $N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$
The rate of reaction is given by: $\text{Rate} = -\frac{d[N_{2}]}{dt} = -\frac{1}{3} \frac{d[H_{2}]}{dt} = \frac{1}{2} \frac{d[NH_{3}]}{dt}$
Given that the rate of formation of $NH_{3}$ is $\frac{d[NH_{3}]}{dt} = 0.088 \ mol \ dm^{-3} \ s^{-1}$.
Substituting this into the rate expression: $-\frac{d[N_{2}]}{dt} = \frac{1}{2} \times \frac{d[NH_{3}]}{dt}$
$-\frac{d[N_{2}]}{dt} = \frac{1}{2} \times 0.088 \ mol \ dm^{-3} \ s^{-1} = 0.044 \ mol \ dm^{-3} \ s^{-1}$
Therefore,the rate of consumption of $N_{2}$ is $0.044 \ mol \ dm^{-3} \ s^{-1}$.

(C) The instantaneous rate of a reaction $aA + bB \longrightarrow cC + dD$ is expressed as:
$-\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$
Comparing this with the given expression,the stoichiometric coefficients for reactants $A$ and $B$ are $a$ and $b$,and for products $C$ and $D$ are $c$ and $d$ respectively.
Therefore,the reaction is $aA + bB \longrightarrow cC + dD$.

(D) For the reaction,$2 NH_{3(g)} \xrightarrow{Pt} N_{2(g)} + 3 H_{2(g)}$
The rate of reaction is given by:
$\text{Rate} = -\frac{1}{2} \frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt}$
Given that the rate of reaction is $2.5 \times 10^{-6} \ mol \ dm^{-3} \ sec^{-1}$,we have:
$\text{Rate} = \frac{1}{3} \frac{d[H_2]}{dt} = 2.5 \times 10^{-6} \ mol \ dm^{-3} \ sec^{-1}$
Therefore,the rate of formation of $H_{2(g)}$ is:
$\frac{d[H_2]}{dt} = 3 \times 2.5 \times 10^{-6} \ mol \ dm^{-3} \ sec^{-1} = 7.5 \times 10^{-6} \ mol \ dm^{-3} \ sec^{-1}$

(B) The rate of reaction is determined by the stoichiometry of the balanced chemical equation.
For the reaction: $CH_3Br_{(aq)} + OH_{(aq)}^{-} \longrightarrow CH_3OH_{(aq)} + Br_{(aq)}^{-}$
The rate expression is: $-\frac{d[CH_3Br]}{dt} = -\frac{d[OH^{-}]}{dt} = \frac{d[CH_3OH]}{dt} = \frac{d[Br^{-}]}{dt}$
Given that the rate of consumption of $OH_{(aq)}^{-}$ is $x \ mol \ dm^{-3} \ s^{-1}$,i.e.,$-\frac{d[OH^{-}]}{dt} = x$.
Since the stoichiometric coefficients of $OH^{-}$ and $Br^{-}$ are both $1$,the rate of formation of $Br_{(aq)}^{-}$ is equal to the rate of consumption of $OH_{(aq)}^{-}$.
Therefore,the rate of formation of $Br_{(aq)}^{-}$ is $x \ mol \ dm^{-3} \ s^{-1}$.

(B) The given reaction is $3 I^{-} + S_2 O_8^{2-} \rightarrow I_3^{-} + 2 SO_4^{2-}$.
According to the rate law expression,the rate of reaction is given by: $\text{Rate} = -\frac{1}{3} \frac{d[I^{-}]}{dt} = \frac{1}{2} \frac{d[SO_4^{2-}]}{dt}$.
We are given $\frac{d[SO_4^{2-}]}{dt} = 2.2 \times 10^{-2} \ mol \ dm^{-3} \ s^{-1}$.
Rearranging the expression to find $-\frac{d[I^{-}]}{dt}$: $-\frac{d[I^{-}]}{dt} = \frac{3}{2} \times \frac{d[SO_4^{2-}]}{dt}$.
Substituting the given value: $-\frac{d[I^{-}]}{dt} = \frac{3}{2} \times 2.2 \times 10^{-2} = 3.3 \times 10^{-2} \ mol \ dm^{-3} \ s^{-1}$.

(C) The rate of reaction is expressed as:
$-\frac{1}{2} \frac{\Delta [N_2O_5]}{\Delta t} = \frac{1}{4} \frac{\Delta [NO_2]}{\Delta t} = \frac{\Delta [O_2]}{\Delta t}$
Given that $-\frac{\Delta [N_2O_5]}{\Delta t} = x \ mol \ dm^{-3} \ s^{-1}$.
Substituting this into the rate expression:
$\frac{1}{2} (x) = \frac{1}{4} \frac{\Delta [NO_2]}{\Delta t}$
$\frac{\Delta [NO_2]}{\Delta t} = 4 \times \frac{x}{2} = 2 x \ mol \ dm^{-3} \ s^{-1}$
Therefore,the average rate of formation of $NO_{2_{(g)}}$ is $2 x \ mol \ dm^{-3} \ s^{-1}$.

(D) The rate of reaction for $2A + B \longrightarrow 3C$ is given by: $\text{Rate} = -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{3} \frac{d[C]}{dt}$.
Given,the rate of appearance of $C$ is $\frac{d[C]}{dt} = 1.3 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
From the rate expression,$-\frac{1}{2} \frac{d[A]}{dt} = \frac{1}{3} \frac{d[C]}{dt}$.
Therefore,the rate of disappearance of $A$ is $-\frac{d[A]}{dt} = \frac{2}{3} \frac{d[C]}{dt}$.
$-\frac{d[A]}{dt} = \frac{2}{3} \times (1.3 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}) = 0.866 \times 10^{-4} \ mol \ L^{-1} \ s^{-1} = 8.66 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$.

(A) For a general reaction $aA + bB \rightarrow cC + dD$,the average rate is given by: $\text{Rate} = -\frac{1}{a} \frac{\Delta[A]}{\Delta t} = -\frac{1}{b} \frac{\Delta[B]}{\Delta t} = \frac{1}{c} \frac{\Delta[C]}{\Delta t} = \frac{1}{d} \frac{\Delta[D]}{\Delta t}$.
For the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,the stoichiometric coefficients are $1$,$3$,and $2$ respectively.
Thus,the rate expression is: $-\frac{\Delta[N_2]}{\Delta t} = -\frac{1}{3} \frac{\Delta[H_2]}{\Delta t} = \frac{1}{2} \frac{\Delta[NH_3]}{\Delta t}$.

(D) The rate of reaction is given by the expression:
$\text{Rate} = -\frac{1}{2} \frac{d[N_2 O_5]}{dt} = +\frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$
We are given that $-\frac{d[N_2 O_5]}{dt} = 0.02 \ mol \ dm^{-3} \ s^{-1}$.
To find the rate of formation of $NO_2$,we use the relation:
$\frac{1}{4} \frac{d[NO_2]}{dt} = -\frac{1}{2} \frac{d[N_2 O_5]}{dt}$
$\frac{d[NO_2]}{dt} = 2 \times \left( -\frac{d[N_2 O_5]}{dt} \right)$
$\frac{d[NO_2]}{dt} = 2 \times 0.02 \ mol \ dm^{-3} \ s^{-1} = 0.04 \ mol \ dm^{-3} \ s^{-1}$

(A) For the reaction $4 NH_{3(g)} + 5 O_{2(g)} \rightarrow 4 NO_{(g)} + 6 H_2O_{(g)}$,the rate expression is given by:
$-\frac{1}{4} \frac{d[NH_3]}{dt} = -\frac{1}{5} \frac{d[O_2]}{dt} = +\frac{1}{4} \frac{d[NO]}{dt} = +\frac{1}{6} \frac{d[H_2O]}{dt}$
Given that the rate of formation of $NO$ is $\frac{d[NO]}{dt} = 3.6 \times 10^{-3} \ mol \ L^{-1} \ sec^{-1}$.
Equating the terms for $NO$ and $H_2O$:
$\frac{1}{4} \frac{d[NO]}{dt} = \frac{1}{6} \frac{d[H_2O]}{dt}$
$\frac{d[H_2O]}{dt} = \frac{6}{4} \times \frac{d[NO]}{dt} = 1.5 \times 3.6 \times 10^{-3} \ mol \ L^{-1} \ sec^{-1}$
$= 5.4 \times 10^{-3} \ mol \ L^{-1} \ sec^{-1}$

(C) The rate of a reaction is defined as the change in concentration of a reactant or product per unit time.
Concentration is typically expressed in $mol \ dm^{-3}$ (or $M$) and time $(t)$ is expressed in seconds $(s)$,minutes $(min)$,or hours $(h)$.
Therefore,the unit for the rate of reaction is $\frac{\text{concentration}}{\text{time}} = \frac{mol \ dm^{-3}}{t} = mol \ dm^{-3} \ t^{-1}$.

(D) The rate expression for the given reaction $2 \ NO_{(g)} + O_{2_{(g)}} \rightarrow 2 \ NO_{2_{(g)}}$ is given by:
$-\frac{1}{2} \frac{d[NO]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt}$.
The rate of consumption of $NO_{(g)}$ is defined as $-\frac{d[NO]}{dt}$.
From the expression,$-\frac{d[NO]}{dt} = \frac{d[NO_2]}{dt}$.
Given that $\frac{d[NO_2]}{dt} = 0.052 \ mol \ dm^{-3} \ s^{-1}$,
Therefore,$-\frac{d[NO]}{dt} = 0.052 \ mol \ dm^{-3} \ s^{-1}$.

(A) For a general reaction $aA + bB \rightarrow cC + dD$,the rate of reaction is given by:
$r = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$
For the reaction $2 N_2 O_{5(g)} \rightarrow 4 NO_{2(g)} + O_{2(g)}$,the rate expression is:
$r = -\frac{1}{2} \frac{d[N_2 O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$
Comparing this with the given options,option $A$ is the correct expression for the rate of reaction.

(C) The rate of reaction is given by the expression: $Rate = -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{2} \frac{d[C]}{dt}$.
Given that the rate of disappearance of $A$ is $-\frac{d[A]}{dt} = 0.076 \ mol \ dm^{-3} \ s^{-1}$.
Substituting this into the relation $-\frac{d[B]}{dt} = \frac{1}{2} \frac{d[A]}{dt}$:
$-\frac{d[B]}{dt} = \frac{1}{2} \times 0.076 \ mol \ dm^{-3} \ s^{-1} = 0.038 \ mol \ dm^{-3} \ s^{-1}$.

(A) The rate expression for the reaction $2 NO_{(g)} + O_{2(g)} \rightarrow 2 NO_{2(g)}$ is given by:
$-\frac{1}{2} \frac{d[NO]}{dt} = -\frac{d[O_2]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt}$
Given that the rate of formation of $NO_2$ is $\frac{d[NO_2]}{dt} = 0.052 \ mol \ dm^{-3} \ s^{-1}$.
Equating the rate of disappearance of $O_2$ to the rate of formation of $NO_2$:
$-\frac{d[O_2]}{dt} = \frac{1}{2} \times \frac{d[NO_2]}{dt}$
$-\frac{d[O_2]}{dt} = \frac{1}{2} \times 0.052 \ mol \ dm^{-3} \ s^{-1} = 0.026 \ mol \ dm^{-3} \ s^{-1}$.

(B) The rate of a chemical reaction is defined as the change in concentration of any of the reactants or products per unit time.
Mathematically,for a reaction $A \rightarrow B$,the rate can be expressed as:
Rate $= -\frac{d[A]}{dt} = \frac{d[B]}{dt}$.
Thus,it is expressed in terms of both the rate of consumption of reactants and the rate of formation of products.

(D) For a general reaction $aA + bB \rightarrow cC + dD$,the rate of reaction is expressed as $-\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$.
Given: $\frac{1}{3} \frac{d[X]}{dt} = -\frac{1}{2} \frac{d[Y]}{dt} = -\frac{d[Z]}{dt}$.
This implies that $X$ is a product (positive sign) with a stoichiometric coefficient of $3$,and $Y$ and $Z$ are reactants (negative sign) with stoichiometric coefficients of $2$ and $1$ respectively.
Thus,the reaction is $2Y + Z \rightarrow 3X$.

(A) The rate of reaction is given by the expression: $-\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
Given that the rate of disappearance of $N_2$ is $-\frac{d[N_2]}{dt} = 2.22 \times 10^{-3} \ mol \ dm^{-3} \ s^{-1}$.
From the stoichiometric relationship,we have: $\frac{d[NH_3]}{dt} = 2 \times (-\frac{d[N_2]}{dt})$.
Substituting the given value: $\frac{d[NH_3]}{dt} = 2 \times (2.22 \times 10^{-3} \ mol \ dm^{-3} \ s^{-1}) = 4.44 \times 10^{-3} \ mol \ dm^{-3} \ s^{-1}$.

(A) For a general reaction $aA + bB \rightarrow cC + dD$,the rate of reaction is given by $r = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$.
Applying this to the reaction $2 A + B \rightarrow C + 3 D$,we get:
$r = -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{d[C]}{dt} = \frac{1}{3} \frac{d[D]}{dt}$.
Comparing this with the given options,both $A$ and $B$ are mathematically equivalent expressions for the rate of reaction.

(D) The rate of reaction is given by the expression: $-\frac{1}{2} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$.
Given that $-\frac{d[NO_2]}{dt} = 1.3 \times 10^{-5} \ mol \ L^{-1} \ sec^{-1}$.
Substituting this value into the expression: $\frac{1}{2} \times (1.3 \times 10^{-5}) = \frac{d[O_2]}{dt}$.
Therefore,$\frac{d[O_2]}{dt} = 0.65 \times 10^{-5} \ mol \ L^{-1} \ sec^{-1} = 6.5 \times 10^{-6} \ mol \ L^{-1} \ sec^{-1}$.

(B) For the reaction $2 \ NO + Cl_2 \rightarrow 2 \ NOCl$,the rate of reaction is given by:
Rate $= -\frac{1}{2} \frac{d[NO]}{dt} = -\frac{d[Cl_2]}{dt} = \frac{1}{2} \frac{d[NOCl]}{dt}$.
Since $NO$ is a reactant,its rate of disappearance is $-\frac{d[NO]}{dt}$.
Since $NOCl$ is a product,its rate of appearance is $\frac{d[NOCl]}{dt}$.
From the rate expression: $-\frac{1}{2} \frac{d[NO]}{dt} = \frac{1}{2} \frac{d[NOCl]}{dt}$.
Therefore,$-\frac{d[NO]}{dt} = \frac{d[NOCl]}{dt}$,or $\frac{d[NO]}{dt} = -\frac{d[NOCl]}{dt}$.

(A) The rate of a reaction $aA + bB \longrightarrow cC$ is expressed as: $-\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt}$.
Given the expression: $-\frac{1}{2} \frac{d[x]}{dt} = -\frac{d[y]}{dt} = \frac{1}{2} \frac{d[z]}{dt}$.
Comparing the coefficients,we get $a = 2$,$b = 1$,and $c = 2$.
Therefore,the balanced chemical equation is $2x + y \longrightarrow 2z$.

(C) The given balanced chemical equation is: $3 I_{(aq)}^{-} + S_2 O_{8_{(aq)}}^{2-} \longrightarrow I_{3_{(aq)}}^{-} + 2 SO_{4_{(aq)}}^{2-}$
According to the rate law expression for the reaction,the rate of reaction is given by:
$Rate = -\frac{1}{3} \frac{d[I^-]}{dt} = -\frac{d[S_2 O_8^{2-}]}{dt} = \frac{d[I_3^-]}{dt} = \frac{1}{2} \frac{d[SO_4^{2-}]}{dt}$
Given that the rate of formation of $SO_4^{2-}$ is $\frac{d[SO_4^{2-}]}{dt} = 0.022 \ mol \ dm^{-3} \ sec^{-1}$.
Equating the rates:
$\frac{d[I_3^-]}{dt} = \frac{1}{2} \frac{d[SO_4^{2-}]}{dt}$
$\frac{d[I_3^-]}{dt} = \frac{1}{2} \times 0.022 \ mol \ dm^{-3} \ sec^{-1} = 0.011 \ mol \ dm^{-3} \ sec^{-1}$

(C) For the reaction: $3 X \rightarrow 2 Y + Z$
The rate of reaction is expressed as: $-\frac{1}{3} \frac{d[X]}{dt} = \frac{d[Z]}{dt}$
Given that the rate of disappearance of $X$ $(-\frac{d[X]}{dt})$ is $0.072 \ mol \ s^{-1}$.
Therefore,the rate of appearance of $Z$ $(\frac{d[Z]}{dt})$ is:
$\frac{d[Z]}{dt} = \frac{1}{3} \times (0.072 \ mol \ s^{-1}) = 0.024 \ mol \ s^{-1}$.

(A) For the reaction $2A + B \rightarrow 3C$,the rate of reaction is given by: $\text{Rate} = -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{3} \frac{d[C]}{dt}$.
Given that the rate of appearance of $C$ is $\frac{d[C]}{dt} = 1.3 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
Equating the rate of disappearance of $B$ to the rate of appearance of $C$: $-\frac{d[B]}{dt} = \frac{1}{3} \frac{d[C]}{dt}$.
$-\frac{d[B]}{dt} = \frac{1}{3} \times 1.3 \times 10^{-4} \ mol \ L^{-1} \ s^{-1} = 4.33 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$.

(B) For the reaction $N_{2(g)} + 3 H_{2(g)} \rightarrow 2 NH_{3(g)}$,the rate of reaction is given by:
Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$
Considering the terms for $N_2$ and $H_2$:
$-\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt}$
Multiplying both sides by $-1$:
$\frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt}$
Rearranging the equation to express $\frac{d[H_2]}{dt}$ in terms of $\frac{d[N_2]}{dt}$:
$\frac{d[H_2]}{dt} = 3 \frac{d[N_2]}{dt}$

(D) For the reaction: $4 NH_{3(g)} + 5 O_{2(g)} \longrightarrow 4 NO_{(g)} + 6 H_2 O_{(g)}$
The rate of reaction is given by the expression:
$Rate = -\frac{1}{4} \frac{d[NH_3]}{dt} = -\frac{1}{5} \frac{d[O_2]}{dt} = \frac{1}{4} \frac{d[NO]}{dt} = \frac{1}{6} \frac{d[H_2O]}{dt}$
Given that the rate of formation of $NO$ is $\frac{d[NO]}{dt} = 3.6 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.
From the rate expression,we have:
$-\frac{1}{4} \frac{d[NH_3]}{dt} = \frac{1}{4} \frac{d[NO]}{dt}$
Therefore,the rate of disappearance of $NH_3$ is:
$-\frac{d[NH_3]}{dt} = \frac{d[NO]}{dt} = 3.6 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.

(B) The given reaction is $2A + B \longrightarrow 2C$.
According to the rate law expression for the reaction:
$-\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{2} \frac{d[C]}{dt}$.
Given that the rate of disappearance of $A$ is $-\frac{d[A]}{dt} = 0.076 \ mol \ s^{-1}$.
Substituting this into the rate expression:
$-\frac{d[B]}{dt} = \frac{1}{2} \times (-\frac{d[A]}{dt}) = \frac{1}{2} \times 0.076 \ mol \ s^{-1} = 0.038 \ mol \ s^{-1}$.
Therefore,the rate of disappearance of $B$ is $0.038 \ mol \ s^{-1}$.

(D) For the reaction $2 N_{2}O_{5(g)} \longrightarrow 4 NO_{2(g)} + O_{2(g)}$,the rate of reaction is expressed as:
$\text{Rate} = -\frac{1}{2} \frac{d[N_{2}O_{5}]}{dt} = \frac{1}{4} \frac{d[NO_{2}]}{dt} = \frac{d[O_{2}]}{dt}$
From the equality $\frac{1}{4} \frac{d[NO_{2}]}{dt} = \frac{d[O_{2}]}{dt}$,we can rearrange the terms to find the ratio of the rates of formation:
$\frac{d[NO_{2}]/dt}{d[O_{2}]/dt} = \frac{4}{1} = 4:1$
Thus,the ratio of the rate of formation of $NO_{2(g)}$ to the rate of formation of $O_{2(g)}$ is $4:1$.

(B) For the elementary reaction: $3 H_{2(g)} + N_{2(g)} \rightarrow 2 NH_{3(g)}$
The rate of reaction is expressed as:
Rate $= -\frac{1}{3} \frac{d[H_{2(g)}]}{dt} = -\frac{d[N_{2(g)}]}{dt} = \frac{1}{2} \frac{d[NH_{3(g)}]}{dt}$
To find the relation between $\frac{d[H_{2(g)}]}{dt}$ and $\frac{d[NH_{3(g)}]}{dt}$,we equate their respective parts:
$-\frac{1}{3} \frac{d[H_{2(g)}]}{dt} = \frac{1}{2} \frac{d[NH_{3(g)}]}{dt}$
Multiplying both sides by $2$:
$-\frac{2}{3} \frac{d[H_{2(g)}]}{dt} = \frac{d[NH_{3(g)}]}{dt}$
Thus,option $(B)$ is correct.

(B) For a general reaction $aA + bB \rightarrow cC + dD$,the average rate of reaction is given by: $\text{Rate} = -\frac{1}{a} \frac{\Delta[A]}{\Delta t} = -\frac{1}{b} \frac{\Delta[B]}{\Delta t} = \frac{1}{c} \frac{\Delta[C]}{\Delta t} = \frac{1}{d} \frac{\Delta[D]}{\Delta t}$.
For the reaction $2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)}$,the rate expression is: $\text{Rate} = -\frac{1}{2} \frac{\Delta[SO_2]}{\Delta t} = -\frac{\Delta[O_2]}{\Delta t} = \frac{1}{2} \frac{\Delta[SO_3]}{\Delta t}$.
Comparing this with the given options,$-\frac{\Delta[O_2]}{\Delta t}$ is a correct representation of the average rate of reaction.

(B) For a general reaction $aA + bB \longrightarrow cC + dD$,the rate of reaction is given by: $\text{Rate} = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$.
For the reaction $H_{2} + I_{2} \longrightarrow 2 HI$,the stoichiometric coefficients are $1, 1,$ and $2$ respectively.
Therefore,the differential rate law is: $\text{Rate} = -\frac{d[H_{2}]}{dt} = -\frac{d[I_{2}]}{dt} = \frac{1}{2} \frac{d[HI]}{dt}$.

(A) The given reaction is $3 I^{-} + S_2 O_8^{2-} \longrightarrow I_3^{-} + 2 SO_4^{2-}$.
According to the rate expression:
$Rate = -\frac{1}{3} \frac{d[I^{-}]}{dt} = -\frac{d[S_2 O_8^{2-}]}{dt} = \frac{d[I_3^{-}]}{dt} = \frac{1}{2} \frac{d[SO_4^{2-}]}{dt}$.
We are given $\frac{d[SO_4^{2-}]}{dt} = 2.2 \times 10^{-2} \ mol \ dm^{-3} \ s^{-1}$.
Equating the terms for $S_2 O_8^{2-}$ and $SO_4^{2-}$:
$-\frac{d[S_2 O_8^{2-}]}{dt} = \frac{1}{2} \frac{d[SO_4^{2-}]}{dt}$.
Therefore,$\frac{d[S_2 O_8^{2-}]}{dt} = -\frac{1}{2} \frac{d[SO_4^{2-}]}{dt}$.
Substituting the value:
$\frac{d[S_2 O_8^{2-}]}{dt} = -\frac{1}{2} \times (2.2 \times 10^{-2}) = -1.1 \times 10^{-2} \ mol \ dm^{-3} \ s^{-1}$.

(A) The balanced chemical equation is $N_{2} + 3H_{2} \longrightarrow 2NH_{3}$.
According to the rate expression for the reaction:
$\text{Rate} = -\frac{d[N_{2}]}{dt} = -\frac{1}{3} \frac{d[H_{2}]}{dt} = \frac{1}{2} \frac{d[NH_{3}]}{dt}$.
Given that the rate of disappearance of $H_{2}$ is $-\frac{d[H_{2}]}{dt} = 0.02 \ M/s$.
We need to find the rate of appearance of $NH_{3}$,which is $\frac{d[NH_{3}]}{dt}$.
From the relation $\frac{1}{2} \frac{d[NH_{3}]}{dt} = \frac{1}{3} \left(-\frac{d[H_{2}]}{dt}\right)$:
$\frac{d[NH_{3}]}{dt} = \frac{2}{3} \times 0.02 \ M/s = 0.0133 \ M/s$.

(D) The given reaction is $4 NH_3 + 5 O_2 \rightarrow 4 NO + 6 H_2 O$.
According to the rate expression:
Rate of reaction $= -\frac{1}{4} \frac{d[NH_3]}{dt} = \frac{1}{6} \frac{d[H_2 O]}{dt}$.
Given that the rate of disappearance of $NH_3$ is $-\frac{d[NH_3]}{dt} = 3.6 \times 10^{-3} \ M/s$.
Substituting this into the rate expression:
$\frac{1}{4} (3.6 \times 10^{-3}) = \frac{1}{6} \frac{d[H_2 O]}{dt}$.
Therefore,the rate of formation of water is $\frac{d[H_2 O]}{dt} = \frac{6}{4} \times 3.6 \times 10^{-3} \ M/s$.
$\frac{d[H_2 O]}{dt} = 1.5 \times 3.6 \times 10^{-3} = 5.4 \times 10^{-3} \ M/s$.

(C) The rate of reaction for $2 SO_{2(g)} + O_{2(g)} \rightarrow 2 SO_{3(g)}$ is expressed as:
$Rate = -\frac{1}{2} \frac{d[SO_2]}{dt} = -\frac{d[O_2]}{dt} = +\frac{1}{2} \frac{d[SO_3]}{dt}$
Comparing the terms for $SO_3$ and $O_2$:
$+\frac{1}{2} \frac{d[SO_3]}{dt} = -\frac{d[O_2]}{dt}$
Multiplying both sides by $2$,we get:
$+\frac{d[SO_3]}{dt} = -2 \frac{d[O_2]}{dt}$
Thus,option $C$ is correct.