Zero order reaction Questions in English

(B) The reactant $A$ decomposes at a constant rate of $10 \%$ per hour.
This indicates that the rate of reaction is independent of the concentration of the reactant.
Therefore,the reaction follows zero-order kinetics.
The general unit for the rate constant of a zero-order reaction is $mol \ L^{-1} \ time^{-1}$.
Given the options,the correct unit is $mol \ L^{-1} \ sec^{-1}$.

(C) For a zero order reaction,the rate law is given by $[A] = [A]_0 - kt$.
At half-life,$[A] = \frac{[A]_0}{2}$,so $\frac{[A]_0}{2} = [A]_0 - kt_{1/2}$.
This simplifies to $t_{1/2} = \frac{[A]_0}{2k}$.
Therefore,the half-life period is directly proportional to the initial concentration $[A]_0$.

(B) The rate of a zero order reaction is defined by the rate law expression $Rate = k[A]^0$,where $k$ is the rate constant and $[A]$ is the concentration of the reactant.
Since any value raised to the power of $0$ is $1$,the rate becomes $Rate = k$.
Therefore,the rate of a zero order reaction is independent of the concentrations of the reactants.

(C) For a reaction of order $n$,the unit of rate constant $k$ is given by $(mol \, L^{-1})^{1-n} \, s^{-1}$.
For a zero order reaction,$n = 0$.
Therefore,the unit of $k = (mol \, L^{-1})^{1-0} \, s^{-1} = mol \, L^{-1} \, s^{-1}$.

(A) The rate of a reaction is defined as the change in concentration of reactants or products per unit time.
For a zero-order reaction,the rate of reaction is independent of the concentration of the reactants,i.e.,$\text{Rate} = k[A]^0 = k$.
In the given data,the rate of the reaction remains approximately constant $(2.8 \times 10^{-2} \ mole \ litre^{-1} \ sec^{-1})$ at all given time intervals $(0, 10, 20, 30 \ sec)$.
Since the rate does not change with time,the reaction must be of zero order.

(C) For a zero order reaction,the rate law is given by $Rate = k[Reactant]^0 = k$.
Since the rate is equal to the rate constant $k$,the rate of the reaction is independent of the concentration of the reactant and remains constant throughout the reaction.

(B) For a zero order reaction,the rate of reaction is independent of the concentration of the reactant.
Therefore,the rate law is expressed as $\frac{dx}{dt} = k[A]^0$,where $k$ is the rate constant and $[A]$ is the concentration of the reactant $A$.

(A) For a reaction of order $n$,the unit of rate constant $k$ is given by $(Conc.)^{1-n} \times (time)^{-1}$.
For a zero order reaction,$n = 0$.
Therefore,the unit is $(Conc.)^{1-0} \times (time)^{-1} = Conc. \times time^{-1}$.

(C) For a zero order reaction,the half-life is given by $t_{1/2} = \frac{[A]_0}{2k}$.
Let the initial concentration be $[A]_0 = a$.
Then,$t_{1/2} = \frac{a}{2k} \dots (1)$.
When the initial concentration is reduced to one-fourth,the new concentration is $[A]_0' = \frac{1}{4}a$.
The new half-life is $t_{1/2}' = \frac{\frac{1}{4}a}{2k} = \frac{1}{4} \times \frac{a}{2k} \dots (2)$.
Comparing equation $(2)$ with $(1)$,we get $t_{1/2}' = \frac{1}{4} t_{1/2}$.
Therefore,the time taken for half of the reaction to complete becomes one-fourth of the original time. Hence,option $C$ is correct.

(D) For a zero order reaction,the rate is independent of the concentration of reactants,so the rate remains constant.
The unit of rate constant for a zero order reaction is $mol \ L^{-1} \ s^{-1}$.
Statement $(a)$ is false because $sec^{-1}$ is the unit for a first order reaction.
Statement $(c)$ is false because the rate of a zero order reaction does not change with the concentration of reactants.
Therefore,both $(a)$ and $(c)$ are incorrect statements.

(B) The rate law for a reaction is given by $Rate = k[A]^n$,where $n$ is the order of the reaction.
If the rate of the reaction is equal to the rate constant $(Rate = k)$,then $[A]^n$ must be equal to $1$.
This is only possible when $n = 0$.
Therefore,the reaction is of zero order.

(A) For a zero order reaction,the rate law is given by: $\text{Rate} = k[A]^0 = k$.
Since $\text{Rate} = \frac{d[Concentration]}{dt}$,the unit of the rate constant $(k)$ is the same as the unit of the rate of reaction.
Therefore,the unit of the velocity constant is $\frac{\text{Concentration}}{\text{Time}} = \text{Concentration} \times \text{Time}^{-1}$.

(B) For a zero order reaction,the rate of reaction is independent of the concentration of reactants.
Mathematically,the rate law is expressed as: $Rate = k[Reactant]^0 = k$.
Thus,the rate is equal to the rate constant $k$,which is independent of the reactant concentration.

(A) . For a zero-order reaction,the rate is independent of the concentration of the reactant,i.e.,$Rate = k[R]^0 = k$.
This leads to the integrated rate law: $[R]_t = [R]_0 - kt$.
The reaction is complete when $[R]_t = 0$,which occurs at time $t = [R]_0 / k$.
Since this is a finite value,zero-order reactions reach completion in a finite time.
For reactions of order $n > 0$,the time required for completion is mathematically infinite.

(D) For a zero-order reaction,the half-life is given by $t_{1/2} = \frac{[R]_0}{2k}$.
Since $t_{1/2} \propto [R]_0$,if the initial concentration $[R]_0$ is doubled,the half-life $t_{1/2}$ also doubles.
Therefore,the reaction is of zero order.

(A) For an $n^{th}$ order reaction,the half-life $t_{1/2}$ is related to the initial concentration $[A]_0$ as $t_{1/2} \propto [A]_0^{1-n}$.
Given that the plot of $t_{1/2}$ versus $[A]_0$ is a straight line,this implies $t_{1/2} \propto [A]_0^1$.
Comparing the powers,$1-n = 1$,which gives $n = 0$.
For a zero-order reaction,$t_{1/2} = \frac{[A]_0}{2k}$.
When $[A]_0 = 2 \ mol \ L^{-1}$,$t_{1/2} = 10 \ min$.
So,$10 = \frac{2}{2k} \implies k = 0.1 \ mol \ L^{-1} \ min^{-1}$.
When $[A]_0 = 4 \ mol \ L^{-1}$,$t = \frac{4}{2 \times 0.1} = \frac{4}{0.2} = 20 \ min$.
Thus,$n = 0$ and $t = 20 \ min$.

(D) For a zero-order reaction,the half-life is given by $t_{1/2} = \frac{[A]_0}{2k}$.
Given $t_{1/2} = 1 \ h$ and $[A]_0 = 2.0 \ mol \ L^{-1}$,we have $1 = \frac{2.0}{2k}$,which gives the rate constant $k = 1.0 \ mol \ L^{-1} \ h^{-1}$.
For a zero-order reaction,the rate law is $[A]_t = [A]_0 - kt$.
To find the time $t$ for the concentration to change from $0.50 \ mol \ L^{-1}$ to $0.25 \ mol \ L^{-1}$,we use $\Delta [A] = k \times t$.
$0.50 - 0.25 = 1.0 \times t$.
$0.25 = 1.0 \times t$.
Therefore,$t = 0.25 \ h$.

(A) The general formula for the half-life of a reaction of order $n$ is $t_{1/2} \propto a^{1-n}$.
Given that $t_{1/2} \propto a^1$,we can equate the exponents: $1 - n = 1$.
Solving for $n$,we get $n = 0$.
Therefore,the reaction is of zero order.

(B) For a zero-order reaction,the rate of the reaction is independent of the concentration of the reactants.
Mathematically,$\text{Rate} = k[A]^0 = k$.
Therefore,the rate of reaction remains constant regardless of the concentration of $A$.

(A) For a zero-order reaction,the integrated rate equation is: $[A] = [A]_0 - kt$.
Given:
Rate constant $k = 0.2 \ mol \ m^{-3} \ h^{-1}$.
Time $t = 30 \ minutes = 0.5 \ h$.
Concentration at time $t$,$[A] = 0.05 \ mol \ m^{-3}$.
Substituting the values into the equation:
$0.05 = [A]_0 - (0.2 \times 0.5)$
$0.05 = [A]_0 - 0.1$
$[A]_0 = 0.05 + 0.1 = 0.15 \ mol \ m^{-3}$.

(A) For a zero-order reaction,the half-life period is given by the formula: $t_{\frac{1}{2}} = \frac{[R]_0}{2k}$.
From this expression,it is clear that $t_{\frac{1}{2}} \propto [R]_0$,where $[R]_0$ is the initial concentration and $k$ is the rate constant.
Therefore,the half-life is directly proportional to the initial concentration of the reactant.

(A) For a zero-order reaction,the half-life is given by $t_{1/2} = \frac{[R]_0}{2k}$.
Given $[R]_0 = 2 \ M$ and $t_{1/2} = 1 \ h$,we find the rate constant $k = \frac{[R]_0}{2t_{1/2}} = \frac{2}{2 \times 1} = 1 \ M \ h^{-1}$.
The time $t$ required for the concentration to change from $[R]_1 = 0.5 \ M$ to $[R]_2 = 0.25 \ M$ is given by $t = \frac{[R]_1 - [R]_2}{k}$.
Substituting the values,$t = \frac{0.5 - 0.25}{1} = 0.25 \ h$.

(B) To determine the order of the reaction,we check the rate of change of concentration with respect to time.
For a zero-order reaction,the rate is constant: $Rate = -\frac{d[R]}{dt} = k$.
Let's calculate the rate for different intervals:
Interval $1$: $t = 0$ to $0.05 \ min$,$\Delta [R] = 1.0 - 0.76 = 0.24 \ M$,$\Delta t = 0.05 \ min$. Rate $= \frac{0.24}{0.05} = 4.8 \ M \ min^{-1}$.
Interval $2$: $t = 0.05$ to $0.12 \ min$,$\Delta [R] = 0.76 - 0.40 = 0.36 \ M$,$\Delta t = 0.07 \ min$. Rate $= \frac{0.36}{0.07} \approx 5.14 \ M \ min^{-1}$.
Interval $3$: $t = 0.12$ to $0.18 \ min$,$\Delta [R] = 0.40 - 0.10 = 0.30 \ M$,$\Delta t = 0.06 \ min$. Rate $= \frac{0.30}{0.06} = 5.0 \ M \ min^{-1}$.
Since the rate of reaction is approximately constant $(4.8, 5.14, 5.0)$,the reaction follows zero-order kinetics.

(A) The rate law for a reaction is given by: $\text{Rate} = k[A]^n$,where $k$ is the rate constant and $n$ is the order of the reaction.
Given that the rate of reaction is equal to the rate constant,we have: $\text{Rate} = k$.
Substituting this into the rate law: $k = k[A]^n$.
This implies $[A]^n = 1$,which is only possible if $n = 0$.
Therefore,the reaction is a zero-order reaction.

(D) For a zero-order reaction,the integrated rate equation is given by:
$Kt = [A]_0 - [A]_t$
Given:
$K = 2 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$
$t = 25 \ s$
$[A]_t = 0.5 \ M$
Substituting the values into the equation:
$(2 \times 10^{-2}) \times 25 = [A]_0 - 0.5$
$0.5 = [A]_0 - 0.5$
$[A]_0 = 0.5 + 0.5 = 1.0 \ M$

(A) The general formula for the unit of the rate constant $K$ for a reaction of order $n$ is given by:
$(mol \ L^{-1})^{1-n} \ s^{-1}$.
For a zero-order reaction,$n = 0$.
Substituting $n = 0$ into the formula:
$(mol \ L^{-1})^{1-0} \ s^{-1} = mol \ L^{-1} \ s^{-1}$.
Therefore,the unit of $K$ for a zero-order reaction is $mol \ L^{-1} \ s^{-1}$.

(C) For a zero-order reaction,the integrated rate equation is given by $t = \frac{[R]_0 - [R]_t}{K}$.
Given that the initial concentration $[R]_0 = a$ and for $100\%$ completion,the final concentration $[R]_t = 0$.
Substituting these values into the equation:
$t = \frac{a - 0}{K} = \frac{a}{K}$.

(A) The adsorption of gases on solid surfaces,such as the decomposition of $NH_3$ on a tungsten surface at high pressure,follows zero-order kinetics. This is because the surface of the catalyst becomes completely covered with gas molecules,and the rate of reaction becomes independent of the concentration of the reactant. Therefore,the order of reaction is $0$.

(C) The reaction is of zero order because the units of the rate constant are $mol \, L^{-1} \, s^{-1}.$
For a zero order reaction,the change in concentration $x$ is given by $x = K \cdot t.$
Given $K = 0.6 \times 10^{-3} \, mol \, L^{-1} \, s^{-1}$ and $t = 20 \, \text{minutes} = 20 \times 60 \, s = 1200 \, s.$
Substituting the values: $x = (0.6 \times 10^{-3}) \times 1200 = 0.72 \, M.$
Thus,the concentration of $B$ formed after $20$ minutes is $0.72 \, M.$

(A) The rate law expression for a zero order reaction is given by: $\text{Rate} = k[A]^0$.
Since the rate of reaction is expressed in units of concentration per unit time $(mol\, L^{-1}\, s^{-1})$,we have:
$mol\, L^{-1}\, s^{-1} = k \times (mol\, L^{-1})^0$.
Since any value raised to the power of $0$ is $1$,the unit of $k$ is equal to the unit of the rate of reaction.
Therefore,the unit of the rate constant for a zero order reaction is $mol\, L^{-1}\, s^{-1}$.

(C) For a zero-order reaction,the half-life is given by $t_{1/2} = \frac{[A]_0}{2k}$.
Given $t_{1/2} = 1 \ hr$ and $[A]_0 = 2.0 \ mol \ L^{-1}$,we calculate the rate constant $k$:
$k = \frac{[A]_0}{2 t_{1/2}} = \frac{2.0 \ mol \ L^{-1}}{2 \times 1 \ hr} = 1 \ mol \ L^{-1} \ hr^{-1}$.
For a zero-order reaction,the time $t$ required for the concentration to change from $[A]_1$ to $[A]_2$ is given by $k = \frac{[A]_1 - [A]_2}{t}$.
Substituting the values: $1 \ mol \ L^{-1} \ hr^{-1} = \frac{0.50 \ mol \ L^{-1} - 0.25 \ mol \ L^{-1}}{t}$.
$t = \frac{0.25 \ mol \ L^{-1}}{1 \ mol \ L^{-1} \ hr^{-1}} = 0.25 \ hr$.

(D) The integrated rate equation for a zero order reaction is given by: $(a-x) = a - kt$.
Comparing this with the equation of a straight line,$y = mx + c$,where $y = (a-x)$,$x = t$,$m = -k$ (slope),and $c = a$ (intercept).
Therefore,a plot of $(a-x)$ $Vs$ $t$ is a straight line with a $-ive$ slope $(-k)$ and a non$-zero$ intercept $(a)$.

(B) The rate of polymerization is zero order: $[A]_t = [A]_0 - Kt$.
Given $[A]_0 = 5 \times 10^{-3} \, moles$,$t = 20 \, minutes$,and $K = 10^{-X} \, moles/minute$.
Remaining moles of $A$ after $20$ minutes: $n_A = 5 \times 10^{-3} - 20 \times 10^{-X}$.
After adding $6 \times 10^{-3} \, moles$ of $C$,the total moles of solute in $5 \, moles$ $(0.005 \, kg)$ of water is $n_{total} = (5 \times 10^{-3} - 20 \times 10^{-X}) + 6 \times 10^{-3} = 11 \times 10^{-3} - 20 \times 10^{-X}$.
Using the depression of freezing point formula: $\Delta T_f = K_f \cdot m$.
$0.186 = 1.86 \cdot \frac{n_{total}}{0.005 \, kg}$ $\Rightarrow 0.1 = \frac{n_{total}}{0.005}$ $\Rightarrow n_{total} = 0.0005 \, moles$.
Wait,recalculating based on molality $m = \frac{n_{total}}{0.005} = 0.1 \Rightarrow n_{total} = 0.1 \times 0.005 = 0.0005 \, moles$.
Setting $11 \times 10^{-3} - 20 \times 10^{-X} = 0.5 \times 10^{-3}$ $\Rightarrow 20 \times 10^{-X} = 10.5 \times 10^{-3} = 1.05 \times 10^{-2}$.
$10^{-X} = 0.0525 \times 10^{-2} = 5.25 \times 10^{-4}$.
Re-evaluating the provided solution logic: $0.1 = \frac{n_{total}}{0.090 \, kg}$ (assuming $5 \, moles$ of water is $0.090 \, kg$ is incorrect,$5 \, moles = 0.090 \, kg$ is $90 \, g$).
Given the provided solution steps: $0.1 = \frac{11 \times 10^{-3} - 20 \times 10^{-X}}{0.090}$ $\Rightarrow 0.009 = 0.011 - 20 \times 10^{-X}$ $\Rightarrow 20 \times 10^{-X} = 0.002$ $\Rightarrow 10^{-X} = 10^{-4}$ $\Rightarrow X = 4$.

(C) For a zero order reaction,the concentration at time $t$ is given by $[A] = [A]_0 - kt$.
The half-life $(t_{1/2})$ is the time when $[A] = \frac{a}{2}$,so $t_{1/2} = \frac{a}{2k}$.
Substituting the values: $t_{1/2} = \frac{1}{2 \times 0.001} = 500 \, sec$.
The completion time $(T)$ is the time when $[A] = 0$,so $T = \frac{a}{k}$.
Substituting the values: $T = \frac{1}{0.001} = 1000 \, sec$.

(B) The rate of reaction is given as $1.60 \times 10^{-2} \ mol \ L^{-1} s^{-1}$ at all time intervals.
Since the rate of reaction is independent of the concentration of the reactant (as it remains constant over time),the reaction follows zero-order kinetics.
Therefore,the order of the reaction is $0$.

(D) For a zero order reaction,the integrated rate equation is given by: $(a - x) = -kt + a$.
Comparing this with the equation of a straight line,$y = mx + c$,where $y = (a - x)$,$x = t$,$m = -k$,and $c = a$.
Since the slope $m = -k$ is negative and the intercept $c = a$ is non-zero (initial concentration),the plot of $(a - x)$ $vs$ time is linear with a negative slope and a non-zero intercept.

(B) The rate of reaction is given by $\frac{dx}{dt}$.
Since the plot of rate versus time $t$ is a straight line parallel to the time axis,the rate is constant and independent of time.
For a zero-order reaction,the rate law is expressed as $\text{Rate} = k \times [A]^0 \times [B]^0 = k$.
Since the rate is constant and equal to the rate constant $k$,the reaction is of zero order.

(A) For a zero order reaction,the integrated rate equation is $[A] = [A]_0 - kt$.
At the completion of the reaction,the final concentration $[A] = 0$ and the initial concentration $[A]_0 = a$.
Substituting these values: $0 = a - kt_{completion}$.
Therefore,$t_{completion} = \frac{a}{k}$.

(C) The unit of rate constant $(M/sec)$ indicates a zero-order reaction.
For a zero-order reaction: $[A]_t = [A]_0 - kt$
Given: $[A]_0 = 1 \, M$,$k = 0.001 \, M/sec$,$t = 10 \times 60 = 600 \, sec$
$[A]_t = 1 - (0.001 \times 600) = 1 - 0.6 = 0.4 \, M$
Concentration of $B$ formed = $[A]_0 - [A]_t = 1 - 0.4 = 0.6 \, M$.

(A) For a zero order reaction,the rate of reaction is independent of the concentration of the reactant.
$1$. The rate law for this reaction is: $\text{Rate} = k[NH_3]^0 = k$.
$2$. Since the rate is equal to the rate constant $(k)$,option $B$ is true.
$3$. At high pressure,the catalyst surface is fully saturated with $NH_3$ molecules. Therefore,increasing the pressure further does not increase the rate of reaction,making option $C$ true.
$4$. Since the rate is constant,the rate of decomposition of ammonia remains constant,making option $D$ true.
$5$. Option $A$ states that the rate depends on the concentration of $NH_3$,which is false for a zero order reaction. Thus,option $A$ is the correct answer.

(D) For a zero order reaction,the integrated rate equation is given by:
$[A]_t = [A]_0 - kt$
Where $[A]_t$ is the concentration at time $t$,$[A]_0$ is the initial concentration,$k$ is the rate constant,and $t$ is the time.
Given:
$k = 2.0 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$
$[A]_t = 0.5 \ M$
$t = 25 \ s$
Substituting the values into the equation:
$0.5 = [A]_0 - (2.0 \times 10^{-2} \times 25)$
$0.5 = [A]_0 - 0.5$
$[A]_0 = 0.5 + 0.5 = 1.0 \ M$

(B) For a reaction of order $n$,the half-life is given by $t_{1/2} \propto (a_0)^{1-n}$.
Taking logarithm on both sides,we get $\log\,t_{1/2} = (1-n)\log\,a_0 + \text{constant}$.
Comparing this with the equation of a straight line $y = mx + c$,the slope $m = 1-n$.
From the given figure,the slope is $\tan(45^{\circ}) = 1$.
Therefore,$1-n = 1$,which gives $n = 0$.

(C) For a zero-order reaction,the integrated rate law is given by $C_0 - C_t = Kt$.
First,calculate the rate constant $K$ using the half-life formula for a zero-order reaction: $t_{1/2} = \frac{C_0}{2K}$.
Given $t_{1/2} = 6 \ h$ and $C_0 = 0.2 \ M$,we have $6 = \frac{0.2}{2K}$,which gives $K = \frac{0.2}{12} = \frac{1}{60} \ M \ h^{-1}$.
Now,for the second condition where initial concentration $C_0 = 0.5 \ M$ and final concentration $C_t = 0.2 \ M$:
Using $C_0 - C_t = Kt$,we get $0.5 - 0.2 = Kt$.
$0.3 = Kt$.
Substituting $K = \frac{1}{60}$,we get $0.3 = \frac{1}{60} \times t$.
$t = 0.3 \times 60 = 18 \ h$.

(A) The unit of the rate constant is $\mu g/year$,which indicates that the reaction is of zero order.
For a zero order reaction,the integrated rate law is $[A] = [A]_0 - kt$.
Here,$[A]_0 = 5 \ \mu g$,$[A] = 2.5 \ \mu g$,and $k = 0.05 \ \mu g/year$.
Substituting these values: $2.5 \ \mu g = 5 \ \mu g - (0.05 \ \mu g/year) \times t$.
$0.05 \ \mu g/year \times t = 2.5 \ \mu g$.
$t = \frac{2.5 \ \mu g}{0.05 \ \mu g/year} = 50 \ years$.

(B) For a zero order reaction,the integrated rate equation is given by: $[R] = [R]_0 - kt$
Where:
$[R]$ is the concentration of the reactant at time $t = 0.5 \ M$
$[R]_0$ is the initial concentration of the reactant
$k$ is the rate constant $= 3 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$
$t$ is the time $= 25 \ s$
Substituting the values into the equation:
$0.5 = [R]_0 - (3 \times 10^{-2} \times 25)$
$0.5 = [R]_0 - 0.75$
$[R]_0 = 0.5 + 0.75$
$[R]_0 = 1.25 \ M$

(A) For a zero order reaction,the rate constant $k$ is given by $k = \frac{[A]_0 - [A]}{t}$.
At half-life $(t = t_{0.5})$,the concentration $[A] = \frac{[A]_0}{2}$.
Substituting these values: $k = \frac{[A]_0 - \frac{[A]_0}{2}}{t_{0.5}} = \frac{[A]_0}{2t_{0.5}}$.
Rearranging for $t_{0.5}$,we get $t_{0.5} = \frac{[A]_0}{2k}$.
Since $k$ is constant,$t_{0.5} \propto [A]_0$ (where $a$ is the initial concentration $[A]_0$).
Therefore,$t_{0.5} \propto a$.

(B) The rate of dissociation is constant: $10\% \text{ per hour}$.
Since the amount of reactant dissociated is directly proportional to time,the reaction follows zero-order kinetics.
For a zero-order reaction,the rate law is $[R] = [R]_0 - Kt$,which implies $x = Kt$,where $x$ is the concentration of reactant dissociated.
The rate constant $K$ has the same units as the rate of reaction,which is $\text{concentration} \times \text{time}^{-1}$.
Therefore,the unit is $mol \ L^{-1} \ hour^{-1}$.

(A) For a zero order reaction,the rate law is given by: $\text{Rate} = k[A]^0 = k$.
$1$. The rate is independent of the concentration of the reactants,so statement $B$ is true.
$2$. The rate constant $k$ follows the Arrhenius equation,$k = Ae^{-E_a/RT}$,meaning it is dependent on temperature. Therefore,the rate of reaction is dependent on temperature. Statement $A$ is false.
$3$. The half-life for a zero order reaction is given by $t_{1/2} = [A]_0 / 2k$,which depends on the initial concentration of the reactant. Statement $C$ is true.
$4$. The unit of the rate constant for a zero order reaction is the same as the rate,which is $mol \ L^{-1} \ s^{-1}$. Statement $D$ is true.

(D) For a zero order reaction,the rate of reaction is independent of the concentration of the reactant.
$Rate = -\frac{d[A]}{dt} = \frac{d[B]}{dt} = k[A]^0 = k$.
$1$. The concentration of product $[B]$ increases linearly with time $t$ $([B] = kt + [B]_0)$.
$2$. The rate of formation of product,$\frac{d[B]}{dt}$,is constant and independent of time $t$.
$3$. The half-life period $t_{1/2}$ is directly proportional to the initial concentration $[A]_0$ $(t_{1/2} = \frac{[A]_0}{2k})$.
$4$. The time for $3/4$ completion,$t_{3/4}$,is also directly proportional to the initial concentration $[A]_0$ $(t_{3/4} = \frac{3[A]_0}{4k})$.
Looking at the provided options:
- Option $A$ shows $[B]$ vs $t$ as a constant line,which is incorrect.
- Option $B$ shows $\frac{d[B]}{dt}$ vs $t$ as a linear increase,which is incorrect.
- Option $C$ shows $t_{1/2}$ vs $[A]_0$ as a constant line,which is incorrect.
- Option $D$ shows $t_{3/4}$ vs $[A]_0$ as a linear graph passing through the origin,which is correct for a zero order reaction.

(C) The reaction is a zero-order reaction because the unit of the rate constant is $M \ s^{-1}.$
The integrated rate equation for a zero-order reaction is $[A]_t = [A]_0 - Kt.$
Given:
$[A]_0 = 1 \ M$
$K = 0.001 \ M \ s^{-1}$
$t = 10 \ minutes = 10 \times 60 \ s = 600 \ s$
Calculating the concentration of $A$ at time $t$:
$[A]_t = 1 - (0.001 \times 600) = 1 - 0.6 = 0.4 \ M.$
Since the stoichiometry of the reaction $A \to B + C$ is $1:1,$ the concentration of $B$ formed is equal to the amount of $A$ consumed:
$[B]_t = [A]_0 - [A]_t = 1 - 0.4 = 0.6 \ M.$
Therefore,the concentrations of $A$ and $B$ are $0.4 \ M$ and $0.6 \ M$ respectively.