Rate of a reaction Questions in English

(A) The rate of reaction is expressed as:
$Rate = -\frac{1}{2} \frac{\Delta[NO_2]}{\Delta t} = \frac{1}{2} \frac{\Delta[NO]}{\Delta t} = \frac{\Delta[O_2]}{\Delta t}$
Given that the rate of disappearance of $NO_2$ is $-\frac{\Delta[NO_2]}{\Delta t} = x \ mol \ dm^{-3} \ s^{-1}$.
Substituting this into the rate expression:
$\frac{\Delta[O_2]}{\Delta t} = \frac{1}{2} \left( -\frac{\Delta[NO_2]}{\Delta t} \right)$
$\frac{\Delta[O_2]}{\Delta t} = \frac{1}{2} \times x = \frac{x}{2} \ mol \ dm^{-3} \ s^{-1}$

(C) For a general reaction $aA + bB \rightarrow cC$,the rate of reaction is given by:
Rate $= -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = +\frac{1}{c} \frac{d[C]}{dt}$.
For the given reaction $3A + 2B \rightarrow 5C$,the coefficients are $a=3$,$b=2$,and $c=5$.
Substituting these values,we get:
Rate $= -\frac{1}{3} \frac{d[A]}{dt} = -\frac{1}{2} \frac{d[B]}{dt} = +\frac{1}{5} \frac{d[C]}{dt}$.
Therefore,option $C$ is correct.

(D) The rate of reaction is given by: $Rate = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt}$.
First,calculate the rate of disappearance of $N_2O_5$: $-\frac{d[N_2O_5]}{dt} = -\frac{1.4 - 2.0}{300} = \frac{0.6}{300} = 2 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$.
Since the rate of production of $NO_2$ is twice the rate of disappearance of $N_2O_5$ (from the stoichiometry $2:4$ or $1:2$): $\frac{d[NO_2]}{dt} = 2 \times (-\frac{d[N_2O_5]}{dt}) = 2 \times 2 \times 10^{-3} = 4 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$.

(B) For the reaction $2 A + 3 B \rightarrow 4 C + D$,the rate expression is given by:
$-\frac{1}{3} \frac{d[B]}{dt} = +\frac{1}{4} \frac{d[C]}{dt}$
Given that the rate of appearance of $C$ is $\frac{d[C]}{dt} = 2.8 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.
The rate of disappearance of $B$ is $-\frac{d[B]}{dt}$.
From the rate expression: $-\frac{d[B]}{dt} = \frac{3}{4} \frac{d[C]}{dt}$
Substituting the given value: $-\frac{d[B]}{dt} = \frac{3}{4} \times (2.8 \times 10^{-3}) \ mol \ L^{-1} \ s^{-1}$.

(B) For the reaction,$2 \ SO_{2} + O_{2} \rightleftharpoons 2 \ SO_{3}$.
The rate of reaction is given by: $\text{Rate} = -\frac{1}{2} \frac{d[SO_{2}]}{dt} = -\frac{d[O_{2}]}{dt} = +\frac{1}{2} \frac{d[SO_{3}]}{dt}$.
Given that the rate of disappearance of $O_{2}$ is $-\frac{d[O_{2}]}{dt} = 2 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
We need to find the rate of appearance of $SO_{3}$,which is $\frac{d[SO_{3}]}{dt}$.
From the rate expression: $-\frac{d[O_{2}]}{dt} = \frac{1}{2} \frac{d[SO_{3}]}{dt}$.
Therefore,$\frac{d[SO_{3}]}{dt} = 2 \times (-\frac{d[O_{2}]}{dt}) = 2 \times (2 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}) = 4 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.

(B) For the given reaction, $3 \,A \rightarrow 2 \,B$.
According to the rate law expression, the rate of reaction is defined as:
Rate $= -\frac{1}{3} \frac{d[A]}{d t} = +\frac{1}{2} \frac{d[B]}{d t}$.
To find the rate of appearance of $B$, which is $+\frac{d[B]}{d t}$, we multiply both sides by $2$:
$+\frac{d[B]}{d t} = -\frac{2}{3} \frac{d[A]}{d t}$.

(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 $\frac{1}{2} A \rightarrow 2 B$,we have $a = \frac{1}{2}$ and $b = 2$.
Substituting these values:
$-\frac{1}{1/2} \frac{d[A]}{dt} = \frac{1}{2} \frac{d[B]}{dt}$
$-2 \frac{d[A]}{dt} = \frac{1}{2} \frac{d[B]}{dt}$
Multiplying both sides by $\frac{1}{2}$:
$-\frac{d[A]}{dt} = \frac{1}{4} \frac{d[B]}{dt}$

(A) For the reaction $A + 2 B \longrightarrow 3 C + 2 D$,the rate of reaction $(r)$ is given by:
$r = -\frac{d[A]}{dt} = -\frac{1}{2} \frac{d[B]}{dt} = \frac{1}{3} \frac{d[C]}{dt} = \frac{1}{2} \frac{d[D]}{dt}$
Given that the rate of disappearance of $B$ is $-\frac{d[B]}{dt} = x \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$.
Substituting this into the rate expression:
$r = -\frac{1}{2} \frac{d[B]}{dt} = \frac{1}{2} (x \times 10^{-2}) = 0.5x \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$.
The rate of appearance of $C$ is $\frac{d[C]}{dt}$.
From the rate expression,$r = \frac{1}{3} \frac{d[C]}{dt}$,so $\frac{d[C]}{dt} = 3r$.
The ratio of the rate of reaction $(r)$ to the rate of appearance of $C$ $(\frac{d[C]}{dt})$ is:
$\frac{r}{\frac{d[C]}{dt}} = \frac{r}{3r} = \frac{1}{3}$.

(A) The average rate of a reaction is defined as the change in concentration of a reactant or product divided by the time interval taken for the change.
For the reaction $A \rightarrow B$,the average rate is given by:
$\text{Average Rate} = -\frac{\Delta[A]}{\Delta t}$
Here,the change in concentration of $A$ is $\Delta[A] = [A]_{final} - [A]_{initial} = y - x$.
The time interval is $\Delta t = 100 \ \min$.
Therefore,the average rate is $-\frac{y - x}{100} = \frac{x - y}{100}$.
Since rate is always positive,we take the magnitude: $\frac{|x - y|}{100} \ mol \ L^{-1} \ \min^{-1}$.

(B) The average rate of reaction is given by the formula: $\text{Average Rate} = -\frac{\Delta[R]}{\Delta t}$.
Given,$\Delta[R] = [R]_2 - [R]_1 = 0.03 - 0.04 = -0.01 \ mol \ L^{-1}$.
Given,$\Delta t = 40 \ min = 40 \times 60 \ s = 2400 \ s$.
Substituting the values: $\text{Average Rate} = -\frac{-0.01}{2400} = \frac{0.01}{2400} \ mol \ L^{-1} \ s^{-1}$.
$\text{Average Rate} = \frac{1 \times 10^{-2}}{24 \times 10^2} = \frac{1}{24} \times 10^{-4} \approx 0.04167 \times 10^{-4} = 4.167 \times 10^{-6} \ mol \ L^{-1} \ s^{-1}$.

(B) For a reaction $A \rightarrow P$,the rate of reaction is defined as the negative of the rate of change of concentration of reactant $A$ with respect to time.
$r_{\text{inst}} = -\frac{d[A]}{dt}$
In the given graph,the $y$-axis represents the concentration of $A$ and the $x$-axis represents time.
The slope of the tangent drawn to the curve at any point gives the value of $\frac{d[A]}{dt}$.
Since the slope of the tangent at point $C$ is $m$,the value of $\frac{d[A]}{dt}$ at point $C$ is $m$.
Therefore,the instantaneous rate of reaction at point $C$ is $-(\text{slope}) = -m$.
However,in the context of magnitude of rate,the instantaneous rate is represented by the absolute value of the slope,which is $m$.

(B) For a reaction $A \rightarrow P$,the rate of reaction is defined as the negative of the rate of change of concentration of reactant $A$ with respect to time,i.e.,$r_{\text{inst}} = -\frac{d[A]}{dt}$.
In the given graph,the $y$-axis represents the concentration of $A$ and the $x$-axis represents time.
The slope of the tangent drawn to the curve at any point $C$ is given by $\frac{d[A]}{dt}$.
Since the curve shows a decrease in concentration with time,the slope $\frac{d[A]}{dt}$ is negative.
Therefore,the instantaneous rate of reaction at point $C$ is equal to the negative of the slope,which is $-(\text{slope})$.
However,in the context of the provided options and the standard interpretation of such graphical problems where $m$ represents the magnitude of the slope (i.e.,$m = |\text{slope}|$),the instantaneous rate is equal to $m$.

(B) The rate of reaction is given by the expression: $-\frac{1}{3} \frac{d[X]}{dt} = \frac{1}{2} \frac{d[Y]}{dt} = \frac{d[Z]}{dt}$
Given that the rate of disappearance of $X$ is $-\frac{d[X]}{dt} = 7.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.
From the rate expression,we have: $\frac{1}{2} \frac{d[Y]}{dt} = -\frac{1}{3} \frac{d[X]}{dt}$
Therefore,the rate of formation of $Y$ is: $\frac{d[Y]}{dt} = \frac{2}{3} \times (-\frac{d[X]}{dt}) = \frac{2}{3} \times 7.2 \times 10^{-3} = 4.8 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.

(A) For the given reaction: $5Br^-{_{\text{(aq)}}} + BrO_3^-{_{\text{(aq)}}} + 6H^+{_{\text{(aq)}}} \rightarrow 3Br_{2\text{(aq)}} + 3H_2O_{\text{(l)}}$
The rate of reaction is defined as the rate of disappearance of reactants divided by their stoichiometric coefficients.
Rate $= -\frac{1}{5} \frac{\Delta[Br^{-}]}{\Delta t} = -\frac{\Delta[BrO_3^{-}]}{\Delta t} = -\frac{1}{6} \frac{\Delta[H^{+}]}{\Delta t} = \frac{1}{3} \frac{\Delta[Br_2]}{\Delta t} = \frac{1}{3} \frac{\Delta[H_2O]}{\Delta t}$
Given that $-\frac{\Delta[Br^{-}]}{\Delta t} = x \ mol \ L^{-1} \ min^{-1}$.
Substituting this into the rate expression:
Rate $= \frac{1}{5} \times (-\frac{\Delta[Br^{-}]}{\Delta t}) = \frac{1}{5} \times x = \frac{x}{5} \ mol \ L^{-1} \ min^{-1}$.

(C) For the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$,the rate of reaction is given by:
Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
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,$-\frac{d[H_2]}{dt} = 3 \times (-\frac{d[N_2]}{dt})$.
$-\frac{d[H_2]}{dt} = 3 \times 0.02 = 0.06 \ mol \ L^{-1} \ s^{-1}$.

(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 $3 \frac{d[NH_3]}{dt} = -2 \frac{d[H_2]}{dt}$.

(D) For the reaction $2 \ A \rightarrow 4 \ B + C$,the rate expression is given by: $-\frac{1}{2} \frac{d[A]}{dt} = \frac{1}{4} \frac{d[B]}{dt}$.
We need to find the rate of disappearance of $A$,which is $-\frac{d[A]}{dt}$.
From the expression: $-\frac{d[A]}{dt} = \frac{2}{4} \frac{d[B]}{dt} = \frac{1}{2} \frac{d[B]}{dt}$.
Given $\frac{\Delta [B]}{\Delta t} = \frac{5 \times 10^{-3} \ mol \ L^{-1}}{10 \ s} = 5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
Therefore,$-\frac{d[A]}{dt} = \frac{1}{2} \times (5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}) = 2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.

(C) For the reaction $2N_2O_{5(g)} \longrightarrow 4NO_{2(g)} + O_{2(g)}$,the rate expression is given by:
$-\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$
Given that the rate of disappearance of $N_2O_5$ is $-\frac{d[N_2O_5]}{dt} = 1.2 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$.
From the rate expression,we have:
$\frac{1}{4} \frac{d[NO_2]}{dt} = \frac{1}{2} \left( -\frac{d[N_2O_5]}{dt} \right)$
$\frac{d[NO_2]}{dt} = 2 \times \left( -\frac{d[N_2O_5]}{dt} \right)$
$\frac{d[NO_2]}{dt} = 2 \times (1.2 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}) = 2.4 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$.

(C) For the reaction $2 A + B \longrightarrow C$,the rate of reaction is given by: $-\frac{1}{2} \frac{d[A]}{d t} = -\frac{d[B]}{d t} = \frac{d[C]}{d t}$.
Given that the rate of formation of $C$ is $\frac{d[C]}{d t} = 2.2 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$.
From the rate expression,we have $-\frac{1}{2} \frac{d[A]}{d t} = \frac{d[C]}{d t}$.
Therefore,$-\frac{d[A]}{d t} = 2 \times \frac{d[C]}{d t}$.
Substituting the value: $-\frac{d[A]}{d t} = 2 \times 2.2 \times 10^{-3} = 4.4 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$.

(C) For the given reaction: $5Br^-{_{\text{(aq)}}} + 6H^+{_{\text{(aq)}}} + BrO_3^-{_{\text{(aq)}}} \rightarrow 3Br_{2\text{(aq)}} + 3H_2O_{\text{(l)}}$
The rate of reaction expression is:
$Rate = -\frac{1}{5} \frac{\Delta[Br^{-}]}{\Delta t} = -\frac{\Delta[BrO_3^{-}]}{\Delta t} = +\frac{1}{3} \frac{\Delta[Br_2]}{\Delta t}$
Given that $-\frac{\Delta[BrO_3^{-}]}{\Delta t} = 0.01 \ mol \ L^{-1} \ min^{-1}$,we equate the terms:
$-\frac{\Delta[BrO_3^{-}]}{\Delta t} = \frac{1}{3} \frac{\Delta[Br_2]}{\Delta t}$
Substituting the given value:
$0.01 = \frac{1}{3} \frac{\Delta[Br_2]}{\Delta t}$
Therefore,$\frac{\Delta[Br_2]}{\Delta t} = 3 \times 0.01 = 0.03 \ mol \ L^{-1} \ min^{-1}$.

(D) For the given reaction: $5Br^-{_{\text{(aq)}}} + BrO_3^-{_{\text{(aq)}}} + 6H^+{_{\text{(aq)}}} \rightarrow 3Br_{2\text{(aq)}} + 3H_2O_{\text{(l)}}$
The rate of reaction is expressed as:
$Rate = -\frac{1}{5} \frac{\Delta[Br^{-}]}{\Delta t} = -\frac{\Delta[BrO_3^{-}]}{\Delta t} = -\frac{1}{6} \frac{\Delta[H^{+}]}{\Delta t} = \frac{1}{3} \frac{\Delta[Br_2]}{\Delta t}$
Given that $-\frac{\Delta[Br^{-}]}{\Delta t} = 0.05 \ mol \ L^{-1} \ min^{-1}$.
Equating the rates:
$-\frac{1}{5} \frac{\Delta[Br^{-}]}{\Delta t} = -\frac{\Delta[BrO_3^{-}]}{\Delta t}$
Substituting the given value:
$-\frac{\Delta[BrO_3^{-}]}{\Delta t} = \frac{1}{5} \times 0.05 = 0.01 \ mol \ L^{-1} \ min^{-1}$.

(D) For the reaction $2 NO_{2(g)} + F_{2(g)} \longrightarrow 2 NO_2F_{(g)}$,the rate of reaction is expressed in terms of the rate of change of partial pressure as:
Rate $= -\frac{1}{2} \left[ \frac{dp(NO_2)}{dt} \right] = -\left[ \frac{dp(F_2)}{dt} \right] = +\frac{1}{2} \left[ \frac{dp(NO_2F)}{dt} \right]$.
Comparing this with the given options,option $D$ is correct.

(B) From the graph,at $t = 0 \ min$,$[A] = 0.05 \ M$ and $[B] = 0 \ M$.
At $t = 10 \ min$,$[A] = 0.04 \ M$ and $[B] = 0.03 \ M$.
The change in concentration of $A$ is $\Delta[A] = 0.05 - 0.04 = 0.01 \ M$.
The change in concentration of $B$ is $\Delta[B] = 0.03 - 0 = 0.03 \ M$.
According to the stoichiometry of the reaction $A \rightarrow nB$,the rate of disappearance of $A$ and appearance of $B$ are related as: $\Delta[B] = n \times \Delta[A]$.
Substituting the values: $0.03 = n \times 0.01$.
Therefore,$n = \frac{0.03}{0.01} = 3$.

(B) For a general chemical reaction $aA + bB \to cC + dD$,the rate of reaction is expressed as:
$\text{rate} = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$.
Comparing the given rate expression with the general form:
$-\frac{1}{6} \frac{d[A]}{dt} = -\frac{1}{a} \frac{d[A]}{dt} \implies a = 6$.
$-\frac{1}{4} \frac{d[B]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} \implies b = 4$.
$\frac{1}{3} \frac{d[C]}{dt} = \frac{1}{c} \frac{d[C]}{dt} \implies c = 3$.
$\frac{1}{4} \frac{d[D]}{dt} = \frac{1}{d} \frac{d[D]}{dt} \implies d = 4$.
Substituting these stoichiometric coefficients into the general reaction equation,we get:
$6A + 4B \to 3C + 4D$.