Zero order reaction Questions in English

(A) For a zero order reaction,the integrated rate equation is: $[A] = [A]_0 - kt$.
At half-life $(t = t_{1/2})$,the concentration of reactant is $[A] = [A]_0 / 2$.
Substituting these values: $\frac{[A]_0}{2} = [A]_0 - k \cdot t_{1/2}$.
Rearranging the terms: $k \cdot t_{1/2} = \frac{[A]_0}{2}$.
Therefore,the rate constant is $k = \frac{[A]_0}{2 \cdot t_{1/2}}$.

(C) For a zero order reaction,the rate of reaction is independent of the concentration of the reactant.
Rate $= -\frac{d[A]}{dt} = k[A]^0 = k$
Integrating this expression:
$-\int d[A] = \int k dt$
$-[A] = kt + C$
At $t = 0$,$[A] = [A]_0$,so $C = -[A]_0$.
Substituting $C$ back into the equation:
$-[A]_t = kt - [A]_0$
$[A]_t = [A]_0 - kt$
Rearranging for $k$:
$kt = [A]_0 - [A]_t$
$k = \frac{[A]_0 - [A]_t}{t}$

(A) The decomposition of $N_{2}O$ on a hot platinum surface is a classic example of a zero order reaction.
Because the catalyst surface becomes saturated with $N_{2}O$ molecules,the reaction rate becomes independent of the concentration of $N_{2}O$.
Therefore,the rate law is $Rate = k[N_{2}O]^0 = k$,which follows zero order kinetics.

(A) For a zero order reaction,the rate law is given by $Rate = k[A]^0 = k$.
Since $Rate = \frac{d[A]}{dt}$,the unit of rate is $mol \ dm^{-3} \ t^{-1}$.
Therefore,the unit of the rate constant $k$ for a zero order reaction is $mol \ dm^{-3} \ t^{-1}$.

(B) zero-order reaction is one where the rate of reaction is independent of the concentration of the reactants.
The decomposition of ammonia $(NH_3)$ on a hot platinum $(Pt)$ surface is a classic example of a zero-order reaction because the metal surface becomes saturated with ammonia molecules,making the rate independent of the concentration of $NH_3$.

(B) For a reaction $aA \rightarrow bB + cC$,the rate of reaction is given by:
$Rate = -\frac{1}{a} \frac{d[A]}{dt} = \frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = K$
Given the reaction $2A \rightarrow B + 3C$,the rate expression is:
$Rate = \frac{1}{3} \frac{d[C]}{dt} = K$
Therefore,the rate of production of $C$ is:
$\frac{d[C]}{dt} = 3 \times K$
$\frac{d[C]}{dt} = 3 \times (3.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}) = 10.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$
Thus,the correct option is $B$.

(B) For a zero order reaction,the integrated rate equation is $[R] = [R]_0 - kt$.
At half-life,$t = t_{1/2}$ and $[R] = \frac{[R]_0}{2}$.
Substituting these values into the equation: $\frac{[R]_0}{2} = [R]_0 - kt_{1/2}$.
Rearranging for $t_{1/2}$: $kt_{1/2} = [R]_0 - \frac{[R]_0}{2} = \frac{[R]_0}{2}$.
Therefore,$t_{1/2} = \frac{[R]_0}{2k}$.
The correct option is $B$.

(D) For a zero order reaction,the integrated rate equation is given by: $[A] = [A]_0 - kt$.
Comparing this with the equation of a straight line,$y = mx + c$,where $y = [A]$,$x = t$,$m = -k$,and $c = [A]_0$.
Thus,the slope of the graph of concentration $[A]$ versus time $t$ is equal to $-k$.

(D) For a zero order reaction,the rate law is given by $Rate = k[R]^0 = k$.
Integrating this,we get $[R]_t = [R]_0 - kt$.
At half-life $(t = t_{1/2})$,the concentration $[R]_t = \frac{[R]_0}{2}$.
Substituting this into the integrated rate equation: $\frac{[R]_0}{2} = [R]_0 - kt_{1/2}$.
Rearranging for $t_{1/2}$,we get $kt_{1/2} = \frac{[R]_0}{2}$,which simplifies to $t_{1/2} = \frac{[R]_0}{2k}$.
Since $k$ is a constant,$t_{1/2} \propto [R]_0$.
Therefore,the correct relation is $t_{1/2} \propto [R]_0$.

(A) For a zero order reaction,the integrated rate equation is $[R] = [R]_0 - kt$.
At the completion of the reaction,the concentration of the reactant $[R]$ becomes $0$.
Substituting this in the equation: $0 = [R]_0 - kt$.
Rearranging for time $t$: $t = \frac{[R]_0}{k}$.
Therefore,the time taken to complete a zero order reaction is $\frac{[R]_0}{k}$.

(D) For a zero-order reaction,the integrated rate equation is given by $[R] = -Kt + [R]_0$.
Rearranging this,we get $Kt = [R]_0 - [R]$,which implies $K = \frac{[R]_0 - [R]}{t}$.
Also,$-K = \frac{[R] - [R]_0}{t}$ is a valid rearrangement of the same equation.
However,the equation $[R]_0 + [R] = -Kt$ is mathematically incorrect as it does not represent the relationship between concentration and time for a zero-order reaction.
Therefore,option $D$ is the incorrect equation.

(C) The decomposition reaction is: $2NH_{3(g)} \rightarrow N_{2(g)} + 3H_{2(g)}$
For a zero-order reaction,the rate of reaction is equal to the rate constant $k$.
$\text{Rate} = k = 2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
The rate of reaction is expressed as: $\text{Rate} = -\frac{1}{2} \frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt}$.
Therefore,the rate of production of $N_2$ is equal to the rate of the reaction.
$\frac{d[N_2]}{dt} = \text{Rate} = 2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.

(B) The balanced chemical equation for the decomposition of ammonia is: $2NH_{3(g)} \rightarrow N_{2(g)} + 3H_{2(g)}$
For a zero-order reaction,the rate of reaction is equal to the rate constant $K$.
Rate $= K = 2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$
From the stoichiometry of the reaction,the rate of reaction is related to the rate of production of $H_2$ as: $\text{Rate} = \frac{1}{3} \frac{d[H_2]}{dt}$
Therefore,$\frac{d[H_2]}{dt} = 3 \times \text{Rate} = 3 \times (2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}) = 7.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$

(D) The unit of the rate constant $k$ is given by $(mol \ L^{-1})^{1-n} \ s^{-1}$,where $n$ is the order of the reaction.
For a zero order reaction $(n = 0)$,the unit is $(mol \ L^{-1})^{1-0} \ s^{-1} = mol \ L^{-1} \ s^{-1}$.
Among the given options,the decomposition of $HI$ on the surface of gold is a zero order reaction because it is a heterogeneous catalytic reaction where the rate is independent of the concentration of the reactant.

(C) For a zero order reaction,the integrated rate equation is $[A] = [A]_0 - kt$.
At $100 \%$ completion,the final concentration $[A] = 0$.
Given initial concentration $[A]_0 = a$.
Substituting these values: $0 = a - kt$.
Therefore,$kt = a$,which gives $t = \frac{a}{k}$.

(C) For a zero order reaction,the integrated rate equation is given by:
$[A] = -kt + [A]_0$
Where $[A]_0$ is the initial concentration $(a)$ and $[A]$ is the concentration at time $t$.
For $100\%$ completion,the remaining concentration $[A] = 0$.
Substituting these values into the equation:
$0 = -kt + a$
$kt = a$
$t = \frac{a}{k}$

(A) For a zero order reaction,the rate law is given by $r = k[A]^0 = k$.
Integrating the rate equation,we get $[A]_t = [A]_0 - kt$.
At half-life $(t = t_{1/2})$,the concentration of reactant $[A]_t = \frac{[A]_0}{2} = \frac{a}{2}$.
Substituting these values: $\frac{a}{2} = a - kt_{1/2}$.
$kt_{1/2} = a - \frac{a}{2} = \frac{a}{2}$.
Therefore,$t_{1/2} = \frac{a}{2k}$.

(A) For a zero order reaction:
Rate $= k[R]^0 = k$. Thus,the rate is independent of concentration. Graph $I$ shows rate vs concentration as a constant line,which represents a zero order reaction.
For the integrated rate law of a zero order reaction:
$[R] = -kt + [R]_0$. This is in the form of $y = mx + c$,where $y = [R]$,$x = t$,$m = -k$,and $c = [R]_0$. Thus,graph $II$ represents a zero order reaction.
Therefore,both $I$ and $II$ represent zero order reactions.

(C) For a zero-order reaction,$A \longrightarrow 3B$,the rate law is $[A] = [A]_0 - kt$.
From the table,at $t = 15 \ s$,$[A] = 0.05 \ mol \ L^{-1}$ and $[A]_0 = 0.1 \ mol \ L^{-1}$.
Substituting these values: $0.05 = 0.1 - k(15) \implies 15k = 0.05 \implies k = \frac{0.05}{15} = \frac{1}{300} \ mol \ L^{-1} \ s^{-1}$.
Now,at $t = 20 \ s$,the concentration of $A$ is $[A] = [A]_0 - kt = 0.1 - (\frac{1}{300}) \times 20 = 0.1 - \frac{20}{300} = 0.1 - 0.0667 = 0.0333 \ mol \ L^{-1}$.
The amount of $A$ reacted is $\Delta[A] = [A]_0 - [A] = 0.1 - 0.0333 = 0.0667 \ mol \ L^{-1}$.
According to the stoichiometry $A \longrightarrow 3B$,the concentration of product $B$ formed is $3 \times \Delta[A] = 3 \times 0.0667 = 0.2001 \ mol \ L^{-1}$.
Rounding to the nearest option,we get $0.198 \ mol \ L^{-1}$ or $1.98 \times 10^{-1} \ mol \ L^{-1}$.

$(A)$ For a zero order reaction:
Rate $= K[A]^0 = K$
Thus, the rate does not change with concentration. Therefore, graph $I$ represents a zero order reaction.
For the integrated rate law of a zero order reaction:
$[R] = -Kt + [R]_0$
Comparing this with $y = mx + c$, the graph of $[R]$ vs $t$ is a straight line with a negative slope equal to $-K$. Thus, graph $II$ also represents a zero order reaction.

(C) For a zero-order reaction $A \longrightarrow 3 B$,the rate law is $[A] = [A_0] - kt$.
From the table,at $t = 0$,$[A_0] = 0.1 \ mol \ L^{-1}$.
At $t = 15 \ s$,$[A] = 0.05 \ mol \ L^{-1}$.
Substituting these values: $0.05 = 0.1 - k(15)$,which gives $k = 0.05 / 15 = 1/300 \ mol \ L^{-1} \ s^{-1}$.
At $t = 20 \ s$,the concentration of $A$ remaining is $[A] = 0.1 - (1/300) \times 20 = 0.1 - 0.0667 = 0.0333 \ mol \ L^{-1}$.
The amount of $A$ reacted is $0.1 - 0.0333 = 0.0667 \ mol \ L^{-1}$.
According to the stoichiometry,$1 \ mol$ of $A$ produces $3 \ mol$ of $B$.
Therefore,concentration of $B = 3 \times 0.0667 = 0.2001 \ mol \ L^{-1} \approx 1.98 \times 10^{-1} \ mol \ L^{-1}$.

(C) For a zero order reaction,the rate of reaction $r$ is equal to the rate constant $k$.
Given $r = k = y \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
The stoichiometric equation is $2 NH_{3(g)} \rightarrow N_{2(g)} + 3 H_{2(g)}$.
The rate of reaction is expressed as $r = -\frac{1}{2} \frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt}$.
Therefore,the rate of formation of hydrogen is $\frac{d[H_2]}{dt} = 3 \times r$.
Substituting the value of $r$,we get $\frac{d[H_2]}{dt} = 3 \times (y \times 10^{-4}) = 3y \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.

(C) For a zero order reaction,the integrated rate equation is given by:
$[R]_t = -kt + [R]_0$
Rearranging for time $t$:
$t = \frac{[R]_0 - [R]_t}{k}$
Given:
$[R]_0 = 0.10 \ M$
$[R]_t = 0.075 \ M$
$k = 2.5 \times 10^{-3} \ M \ s^{-1}$
Substituting the values:
$t = \frac{0.10 - 0.075}{2.5 \times 10^{-3}}$
$t = \frac{0.025}{2.5 \times 10^{-3}}$
$t = 0.01 \times 10^{3} = 10 \ s$

(A) For a zero-order reaction,the rate law is given by: $dx/dt = k[A]^0 = k$.
This means the rate of reaction is independent of the concentration of the reactant.
Therefore,the plot of $dx/dt$ versus $(a-x)$ should be a horizontal line parallel to the $(a-x)$ axis.
However,looking at the standard integrated rate equation for a zero-order reaction: $[A] = [A]_0 - kt$,where $[A] = (a-x)$ and $[A]_0 = a$.
This gives $(a-x) = a - kt$,or $(a-x) = -kt + a$.
Comparing this with the equation of a straight line $y = mx + c$,a plot of $(a-x)$ versus $t$ is a straight line with a negative slope $(-k)$ and an intercept $(a)$.
Among the given options,the plot of $(a-x)$ versus $t$ (which is represented by option $A$ where $a$ is the initial concentration and the y-axis represents the concentration at time $t$) is the correct representation for a zero-order reaction.

(A) For a zero order reaction,the integrated rate equation is given by:
$[A]_t = [A]_0 - kt$
Given:
Rate constant,$k = 1 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$
Initial concentration,$[A]_0 = 0.1 \ mol \ L^{-1}$
Time,$t = 10 \ s$
Substituting the values into the equation:
$[A]_t = 0.1 - (1 \times 10^{-3} \times 10)$
$[A]_t = 0.1 - 10^{-2}$
$[A]_t = 0.1 - 0.01$
$[A]_t = 0.09 \ mol \ L^{-1}$

(C) For a zero order reaction,the integrated rate equation is given by:
$k = \frac{[R]_0 - [R]_t}{t}$
Where:
$k = 0.0030 \ mol \ L^{-1} \ s^{-1}$ (rate constant)
$[R]_0 = 0.10 \ M$ (initial concentration)
$[R]_t = 0.075 \ M$ (concentration at time $t$)
Rearranging the formula to solve for time $t$:
$t = \frac{[R]_0 - [R]_t}{k}$
$t = \frac{0.10 \ M - 0.075 \ M}{0.0030 \ mol \ L^{-1} \ s^{-1}}$
$t = \frac{0.025}{0.0030} \ s = 8.33 \ s$

(D) For a zero order reaction:
Rate $= K[A]^0 = K$
Integrating the rate law,$[A]_t = [A]_0 - Kt$.
At half-life $(t = t_{1/2})$,the concentration $[A]_t = [A]_0/2$.
Substituting these values: $[A]_0/2 = [A]_0 - K t_{1/2}$.
Rearranging gives $K t_{1/2} = [A]_0/2$,or $t_{1/2} = \frac{[A]_0}{2K}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = t_{1/2}$,$x = [A]_0$,and $m = 1/(2K)$,the plot of $t_{1/2}$ versus $[A]_0$ is a straight line passing through the origin with a slope of $1/(2K)$.

(D) The rate law for the reaction is given by $r = k[A]^n$,where $n$ is the order of the reaction.
If the concentration of $A$ is doubled,the new rate $r'$ is $r' = k[2A]^n$.
Given that the rate remains unchanged,$r = r'$,so $k[A]^n = k[2A]^n$.
Dividing both sides by $k[A]^n$,we get $1 = 2^n$.
Since $2^0 = 1$,it follows that $n = 0$.
Therefore,the reaction is a zero-order reaction.

(D) For a zero-order reaction,the integrated rate law is given by $[A] = -kt + [A]_0$.
This equation is in the form of a linear equation $y = mx + c$,where $y = [A]$,$x = t$,$m = -k$ (slope),and $c = [A]_0$ ($y$-intercept).
Therefore,a plot of $[A]$ versus time $(t)$ results in a straight line for a zero-order reaction.
Hence,the correct option is $(D)$.

(A) For a zero order reaction,the integrated rate equation is given by: $[A] = -kt + [A]_0$.
Comparing this with the equation of a straight line,$y = mx + c$,where $y = [A]$,$x = t$,$m = -k$,and $c = [A]_0$.
The slope of the plot is $-k$.
Given,slope $= -3 \times 10^{-3} \ M \ min^{-1}$.
Therefore,$-k = -3 \times 10^{-3} \ M \ min^{-1}$,which implies $k = 3 \times 10^{-3} \ M \ min^{-1}$.

$(A)$ The rate law for a reaction is given by: $r = k[conc.]^n$, where $n$ is the order of reaction.
Rearranging for the rate constant $k$: $k = \frac{r}{[conc.]^n} = \frac{mol \,L^{-1} \,s^{-1}}{(mol \,L^{-1})^n} = mol^{1-n} \,L^{n-1} \,s^{-1}$.
For a zero order reaction, $n = 0$.
Substituting $n = 0$ into the expression: $k = mol^{1-0} \,L^{0-1} \,s^{-1} = mol \,L^{-1} \,s^{-1}$.

(A) For a zero order reaction,the rate constant $k$ is given by the formula $k = \frac{[A]_0 - [A]_t}{t}$.
First,calculate $k$ using the half-life formula $t_{1/2} = \frac{[A]_0}{2k}$.
Given $t_{1/2} = 0.5 \ hr$ and $[A]_0 = 4 \ mol \ L^{-1}$,we have $0.5 = \frac{4}{2k}$,which gives $k = \frac{4}{1} = 4 \ mol \ L^{-1} \ hr^{-1}$.
Now,to find the time $t$ for the concentration to change from $[A]_1 = 2.0 \ mol \ L^{-1}$ to $[A]_2 = 1.0 \ mol \ L^{-1}$,use the integrated rate law: $t = \frac{[A]_1 - [A]_2}{k}$.
Substituting the values: $t = \frac{2.0 - 1.0}{4} = \frac{1.0}{4} = 0.25 \ hr$.
Thus,$t = 1/4 \ hr$.

(B) For a zero order reaction,the rate law is given by: $Rate = k[A]^0 = k$.
Integrating this with respect to time,we get the integrated rate equation: $[A] = -kt + [A]_0$.
This equation follows the form of a straight line equation $y = mx + c$,where $y = [A]$,$x = t$,$m = -k$ (slope),and $c = [A]_0$ (intercept).
Therefore,the plot of concentration $[A]$ versus time $t$ is a straight line with a negative slope equal to $-k$ and a non-zero positive intercept equal to $[A]_0$ on the concentration axis.

(C) For a zero-order reaction,the integrated rate equation is given as:
$k = \frac{[R_0] - [R]}{t}$
where $k$ is the rate constant,$[R_0]$ is the initial concentration,and $[R]$ is the concentration at time $t$.
At half-life time,$t = t_{1/2}$,the concentration of the reactant is $[R] = \frac{[R_0]}{2}$.
Substituting these values into the rate equation:
$k = \frac{[R_0] - \frac{[R_0]}{2}}{t_{1/2}}$
Thus,the correct expression is $k = \frac{[R_0] - \frac{1}{2}[R_0]}{t_{1/2}}$.
Hence,option $(C)$ is the correct answer.

(B) For a zero order reaction,the rate is constant and independent of the concentration of the reactant. Thus,$\frac{dx}{dt} = k$ (constant),which matches graph $(iii)$.
Also,the half-life period is given by $t_{1/2} = \frac{a}{2k}$,which shows that $t_{1/2}$ is directly proportional to the initial concentration $a$. This matches graph $(i)$.
Graph $(ii)$ represents a second order reaction ($1/x$ vs $t$ is linear for second order).
Graph $(iv)$ represents a first order reaction ($log(a-x)$ vs $t$ is linear for first order).
Therefore,graphs $(i)$ and $(iii)$ represent a zero order reaction.

(B) For a zero order reaction:
$1$. The integrated rate law is $[A] = [A]_0 - kt$. Thus,a plot of $[A]$ versus $t$ is a straight line with a negative slope $(-k)$,which matches graph $(i)$.
$2$. The half-life is $t_{1/2} = \frac{[A]_0}{2k}$. Thus,a plot of $t_{1/2}$ versus $[A]_0$ is a straight line passing through the origin with a slope of $\frac{1}{2k}$,which matches graph $(ii)$.
$3$. The rate law is $\text{Rate} = k[A]^0 = k$. The rate is independent of the reactant concentration. Thus,a plot of $\text{Rate}$ versus $[A]$ is a horizontal line,which matches graph $(iv)$.
Graph $(iii)$ shows $\text{Rate} \propto [A]$,which is characteristic of a first order reaction.
Therefore,the correct representations are $(i), (ii),$ and $(iv)$.

(D) The reaction $H_2 + Cl_2 \xrightarrow{hv} 2 \ HCl$ is a photochemical reaction.
Photochemical reactions involving the absorption of light are typically zero order reactions because the rate depends on the intensity of light absorbed rather than the concentration of the reactants.

(B) The graph shows a linear plot of $[AB]$ versus $time$,which indicates a zero-order reaction.
For a zero-order reaction,the rate law is $[AB]_0 - [AB]_t = kt$.
From the graph,at $t = 0 \ s$,$[AB]_0 = 0.6 \ M$.
At $t = 100 \ s$,$[AB]_t = 0.55 \ M$.
Substituting these values: $0.6 - 0.55 = k(100) \implies 0.05 = 100k \implies k = 5 \times 10^{-4} \ M \ s^{-1}$.
The half-life for a zero-order reaction is given by $t_{1/2} = \frac{[AB]_0}{2k}$.
$t_{1/2} = \frac{0.6}{2 \times 5 \times 10^{-4}} = \frac{0.6}{10^{-3}} = 600 \ s$.
Converting to minutes: $t_{1/2} = \frac{600}{60} = 10 \ min$.
Therefore,$x = 10$.

(A) For a zero-order reaction, the rate constant $k$ is related to the half-life $t_{1/2}$ by the formula $t_{1/2} = \frac{[A]_0}{2k}$.
Given $t_{1/2} = 10 \text{ min}$, we can find the initial concentration in terms of $k$: $[A]_0 = 2k \times t_{1/2} = 2k \times 10 = 20k$.
The time required to complete the reaction $(100\%)$ is given by the formula $t_{100\%} = \frac{[A]_0}{k}$.
Substituting the value of $[A]_0$, we get $t_{100\%} = \frac{20k}{k} = 20 \text{ min}$.
Therefore, the reaction completes in $20 \text{ minutes}$.

(D) $1$. The unit of the rate constant is $\text{mol L}^{-1} \text{ s}^{-1}$,which is the characteristic unit for a zero-order reaction.
$2$. For a zero-order reaction,the rate is given by $\text{Rate} = k[X]^0 = k$. Thus,the rate is independent of the concentration of the reactant $X$.
$3$. Statement $A$ is false because the rate remains constant regardless of the concentration of $X$.
$4$. Statement $B$ is false because the reaction is zero-order,not second-order.
$5$. Statement $C$ is false because for a zero-order reaction,the half-life $t_{1/2} = \frac{[R]_0}{2k}$,which depends on the initial concentration.
$6$. Statement $D$ is false because the decomposition of $N_2O_5$ is a first-order reaction.
$7$. Statement $E$ is false because the plot of $\ln \frac{[R]_0}{[R]}$ vs time is linear only for first-order reactions. For zero-order,$[R]$ vs time is linear.
$8$. Since none of the statements $A, B, C, D,$ or $E$ are true,the correct choice is 'None of the above'.

1. The plot is linear, $t_{1/2} \propto [A]_0$, characteristic of zero-order reaction. $t_{1/2} = [A]_0 / 2k$. 2. Slope = $240 / (4 \times 10^{-3}) = 60000$. 3. $x$ corresponds to $t_{1/2}$ at $1.5 \times 10^{-3}$. $x = 60000 \times 1.5 \times 10^{-3} = 90$ s. Closest integer 90.

(D) For a zero-order reaction,the rate of reaction is independent of the concentration of the reactant.
The integrated rate equation for a zero-order reaction is given by $[R]_t = -kt + [R]_0$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = [R]_t$,$x = t$,$m = -k$ (slope),and $c = [R]_0$ (intercept).
$A$ linear plot of concentration $[R]$ vs time with a constant negative slope signifies a zero-order reaction.