First Order reaction Questions in English

(B) For a first-order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{K}$.
Given $t_{1/2} = 10 \text{ minutes}$,the rate constant $K = \frac{0.693}{10} = 0.0693 \text{ min}^{-1}$.
The integrated rate law for a first-order reaction is $t = \frac{2.303}{K} \log \frac{[A]_0}{[A]}$.
Here,$[A] = 10 \% [A]_0 = 0.1 [A]_0$,so $\frac{[A]_0}{[A]} = 10$.
Substituting the values: $t = \frac{2.303}{0.0693} \log(10) = \frac{2.303}{0.0693} \times 1 \approx 33.2 \text{ minutes}$.

(A) For a first-$order$ reaction,the rate equation is $R = R_0 e^{-kt}$.
Taking the natural logarithm on both sides: $\ln R = \ln R_0 - kt$.
This can be rearranged as $\ln(R_0/R) = kt$.
Converting to base $10$: $\log(R_0/R) = \frac{kt}{2.303}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log(R_0/R)$ and $x = t$,the slope is $m = \frac{k}{2.303}$,which is positive.
Therefore,plotting $\log(R_0/R)$ vs time gives a straight line with a positive slope.

(A) For a first-order reaction: $[R] = [R]_0 e^{-kt}$.
Taking natural logarithm on both sides: $\ln [R] = \ln [R]_0 - kt$.
Rearranging the terms: $\ln [R]_0 - \ln [R] = kt$,which gives $\ln \frac{[R]_0}{[R]} = kt$.
Converting to base $10$ logarithm: $2.303 \log \frac{[R]_0}{[R]} = kt$.
Therefore,$\log \frac{[R]_0}{[R]} = \frac{k}{2.303} t$.
Comparing this with the equation of a straight line $y = mx + c$,a plot of $\log \frac{[R]_0}{[R]}$ vs $t$ gives a straight line passing through the origin with a positive slope $m = \frac{k}{2.303}$.

(D) The half-life of a reaction is the time required for the reactant concentration to decrease to one-half its initial value.
For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
From the graph,at $t = 30 \ minutes$,the concentration of reactant $[A]_t$ equals the concentration of product $[B_2]$.
Reaction: $A_5 \rightarrow 5 B_2$
Initial: $a \quad 0$
At $t = 30 \ min$: $a-x \quad 5x$
Given $a-x = 5x$,so $a = 6x$.
Substituting into the rate equation:
$k = \frac{2.303}{30} \log \frac{a}{a-x} = \frac{2.303}{30} \log \frac{6x}{x} = \frac{2.303}{30} \log 6 \approx \frac{2.303 \times 0.778}{30} \approx 0.0597 \ min^{-1}$.
The half-life is $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0597} \approx 116 \ minutes$.
Given the options,$114 \ minutes$ is the closest value.

(A) For a first order reaction,$k = \frac{2.303}{t} \log \frac{a}{a-x}$.
For $t_{1/4}$,$x = \frac{a}{4}$,so $t_{1/4} = \frac{2.303}{k} \log \frac{a}{3a/4} = \frac{2.303}{k} \log \frac{4}{3}$.
For $t_{1/10}$,$x = \frac{a}{10}$,so $t_{1/10} = \frac{2.303}{k} \log \frac{a}{9a/10} = \frac{2.303}{k} \log \frac{10}{9}$.
Taking the ratio: $\frac{t_{1/4}}{t_{1/10}} = \frac{\log(4/3)}{\log(10/9)} = \frac{\log 4 - \log 3}{\log 10 - \log 9} = \frac{2 \log 2 - \log 3}{1 - 2 \log 3}$.
Using $\log 2 = 0.3$ and $\log 3 = 0.477$: $\frac{t_{1/4}}{t_{1/10}} = \frac{2(0.3) - 0.477}{1 - 2(0.477)} = \frac{0.6 - 0.477}{1 - 0.954} = \frac{0.123}{0.046} \approx 2.67$.
Thus,$\frac{t_{1/4}}{t_{1/10}} \times 100 \approx 267$. The closest option is $272$.

(C) For a given chemical reaction,the rate constant $k$ depends only on the temperature and the nature of the reactant.
It is independent of the initial concentration of the reactant.
Since the temperature remains constant at $300 \ K$,the rate constant $k$ will remain the same regardless of the change in initial concentration $[A]$.
Therefore,the rate constant remains $0.125 \ min^{-1}$.

(C) For a first order reaction,the rate equation is given by:
$t = \frac{2.303}{k} \log \frac{a}{a-x}$
Given that the reaction is $90 \%$ complete,we have $a = 100$ and $(a-x) = 100 - 90 = 10$.
The rate constant $k = 6.93 \times 10^{-2} \ min^{-1}$.
Substituting these values into the equation:
$t = \frac{2.303}{6.93 \times 10^{-2}} \log \frac{100}{10}$
$t = \frac{2.303}{0.0693} \times \log(10)$
Since $\log(10) = 1$,we get:
$t = \frac{2.303}{0.0693} \approx 33.23 \ min$.
The nearest integer value is $33 \ min$.

(B) The condition that the half-life $(t_{1/2})$ is independent of the initial concentration $([R]_0)$ is a characteristic property of a first-order reaction.
For a first-order reaction,the integrated rate equation is $\ln[R] = -kt + \ln[R]_0$,which can also be written as $\log[R] = -\frac{kt}{2.303} + \log[R]_0$ or $\log\frac{[R]_0}{[R]} = \frac{kt}{2.303}$.
Graph $A$ ($\ln[R]$ vs $\text{time}$) is a straight line with slope $-k$,which is correct.
Graph $B$ ($[R]$ vs $\text{time}$) represents an exponential decay curve,not a straight line with a constant negative slope. Therefore,this graph is incorrect for a first-order reaction.
Graph $C$ ($\log[R]$ vs $\text{time}$) is a straight line with slope $-\frac{k}{2.303}$,which is correct.
Graph $D$ ($\log\frac{[R]_0}{[R]}$ vs $\text{time}$) is a straight line passing through the origin with slope $\frac{k}{2.303}$,which is correct.

(B) For the reaction $A_{(g)} \rightarrow B_{(g)} + C_{(g)}$,let the initial pressure of $A$ be $P_0 = 200 \ mm$. At time $t = 20 \ min$,let the pressure of $A$ decrease by $x$. The total pressure $P_t = (P_0 - x) + x + x = P_0 + x$. Given $P_t = 250 \ mm$,so $250 = 200 + x$,which gives $x = 50 \ mm$. The pressure of $A$ at $t = 20 \ min$ is $P_A = P_0 - x = 200 - 50 = 150 \ mm$. The rate constant $k = \frac{2.303}{t} \log(\frac{P_0}{P_A}) = \frac{2.303}{20} \log(\frac{200}{150}) = \frac{2.303}{20} \log(\frac{4}{3}) = \frac{2.303}{20} (\log 4 - \log 3) = \frac{2.303}{20} (0.60 - 0.48) = \frac{2.303 \times 0.12}{20} = 0.013818 \ min^{-1}$. The half-life $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.013818} \approx 50.15 \ min$. Thus,the half-life is approximately $50.2 \ min$.

(A) For a first-order reaction,the integrated rate equation is given by: $\ln [R] = -kt + \ln [R]_0$.
Comparing this with the equation of a straight line,$y = mx + c$,where $y = \ln [R]$,$x = t$,$m = -k$ (slope),and $c = \ln [R]_0$ (intercept).
Therefore,the intercept on the $y$-axis is equal to $\ln [R]_0$.

(C) For a first order reaction,the integrated rate equation is given by:
$\log \frac{a}{a-x} = \frac{kt}{2.303}$
Comparing this with the equation of a straight line $y = mx + c$:
Here,$y = \log \frac{a}{a-x}$,$x = t$,and the slope $m = \frac{k}{2.303}$.
Given that the slope $m = 2 \times 10^{-3} \ min^{-1}$,we have:
$\frac{k}{2.303} = 2 \times 10^{-3} \ min^{-1}$
Therefore,$k = 2.303 \times 2 \times 10^{-3} \ min^{-1} = 4.606 \times 10^{-3} \ min^{-1}$.

(C) The decomposition reaction is $O_{3(g)} \longrightarrow O_{2(g)} + O_{(g)}$.
Initial pressure: $100 \ atm, \ 0, \ 0$.
Pressure after time $t$: $(100-x), \ x, \ x$.
Given: $k = 1.0 \times 10^{-3} \ s^{-1}$,$t = 38.38 \ min = 38.38 \times 60 \ s = 2302.8 \ s$.
Using the first-order rate equation: $k = \frac{2.303}{t} \log \frac{p_i}{p_f}$.
$1.0 \times 10^{-3} = \frac{2.303}{2302.8} \log \frac{100}{100-x}$.
$\log \frac{100}{100-x} = \frac{1.0 \times 10^{-3} \times 2302.8}{2.303} \approx 1$.
$\frac{100}{100-x} = 10^1 = 10$.
$100 = 1000 - 10x \implies 10x = 900 \implies x = 90 \ atm$.
Partial pressure of $O_3 = 100 - 90 = 10 \ atm$.
Partial pressure of $O_2 = 90 \ atm$.
Partial pressure of $O = 90 \ atm$.
Thus,the correct option is $C$.

(B) Given,rate constant $(k)$ for first order $= 0.2303 \ min^{-1}$.
For a first order reaction,the time $(t)$ is given by:
$t = \frac{2.303}{k} \log \left( \frac{a}{a-x} \right)$
where $a$ is the initial concentration and $x$ is the amount reacted.
Let the initial concentration $(a) = 1$.
Given that the fraction completed is $9/10$,so $x = 0.9$.
The remaining concentration $(a-x) = 1 - 0.9 = 0.1$.
Substituting the values into the equation:
$t = \frac{2.303}{0.2303} \log \left( \frac{1}{0.1} \right)$
$t = 10 \log(10)$
Since $\log(10) = 1$,we get $t = 10 \times 1 = 10 \ min$.
Therefore,the correct option is $(b)$.

(B) The unit of the rate constant $k$ is $s^{-1}$, which indicates that the reaction is a first-order reaction.
For a first-order reaction, the rate constant equation is given by:
$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given:
$k = 1.15 \times 10^{-3} \,s^{-1}$
$[A]_0 = 6 \,g$
$[A]_t = 3 \,g$
Substituting the values:
$1.15 \times 10^{-3} = \frac{2.303}{t} \log \frac{6}{3}$
$1.15 \times 10^{-3} = \frac{2.303}{t} \log 2$
Using $\log 2 = 0.301$:
$t = \frac{2.303 \times 0.301}{1.15 \times 10^{-3}}$
$t = \frac{0.6932}{1.15 \times 10^{-3}}$
$t \approx 602.8 \,s \approx 603 \,s$

(D) For a first order reaction,the integrated rate equation is $\log(a-x) = -\frac{kt}{2.303} + \log a$. Thus,a plot of $\log(a-x)$ vs $t$ is a straight line with a negative slope,as shown in $(i)$.
For a first order reaction,the half-life is $t_{1/2} = \frac{0.693}{k}$,which is independent of the initial concentration $a$. Thus,a plot of $t_{1/2}$ vs $a$ is a horizontal line,as shown in $(ii)$.
For a first order reaction,the rate is $\frac{dx}{dt} = k(a-x)$. Thus,a plot of $\frac{dx}{dt}$ vs $(a-x)$ is a straight line passing through the origin,as shown in $(iii)$.
Therefore,graphs $(i), (ii),$ and $(iii)$ represent a first order reaction.

(C) Since the concentration of the reactant decreases from $0.6 \ M$ to $0.3 \ M$ (i.e.,halved) in $15 \ min$,the half-life $(t_{1/2})$ for this reaction is $15 \ min$.
For a first-order reaction,the concentration decreases by half in each successive half-life interval.
Starting from $0.1 \ M$:
$0.1 \ M$ $\xrightarrow{15 \ min} 0.05 \ M$ $\xrightarrow{15 \ min} 0.025 \ M$.
Total time taken $= 15 \ min + 15 \ min = 30 \ min$.

(C) The integrated rate equation for a first order reaction is given by:
$k = \frac{2.303}{t} \log \frac{a}{(a-x)}$
Rearranging this equation:
$\frac{kt}{2.303} = \log a - \log (a-x)$
$\log (a-x) = -\frac{k}{2.303} t + \log a$
This equation is in the form of a straight line $y = mx + c$,where $y = \log (a-x)$,$x = t$,slope $m = -\frac{k}{2.303}$,and intercept $c = \log a$.
Therefore,a plot of $\log (a-x)$ versus $t$ yields a straight line with a negative slope.

(C) For a first order reaction,the time $t$ is given by $t = \frac{2.303}{k} \log_{10} \frac{a}{a-x}$.
Let the initial concentration $a = 100$.
For $75 \%$ completion,$x = 75$,so the remaining concentration is $100 - 75 = 25$. Thus,$t_{75\%} = \frac{2.303}{k} \log \frac{100}{25} = \frac{2.303}{k} \log 4$.
For $25 \%$ completion,$x = 25$,so the remaining concentration is $100 - 25 = 75$. Thus,$t_{25\%} = \frac{2.303}{k} \log \frac{100}{75} = \frac{2.303}{k} \log \frac{4}{3}$.
The ratio is $\frac{t_{75\%}}{t_{25\%}} = \frac{\log 4}{\log (4/3)} = \frac{\log 4}{\log 4 - \log 3}$.
Using $\log 4 \approx 0.6020$ and $\log 3 \approx 0.4771$,we get $\frac{0.6020}{0.6020 - 0.4771} = \frac{0.6020}{0.1249} \approx 4.82$.

(C) The unit of the rate constant $k$ is $s^{-1}$,which indicates that the reaction is a $1^{st}$ order reaction.
For a $1^{st}$ order reaction,the integrated rate equation is given by $\ln [A] = \ln [A]_0 - kt$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln [A]$,$x = t$,$m = -k$ (slope),and $c = \ln [A]_0$ (intercept).
Therefore,a plot of $\ln [A]$ versus $t$ will be a straight line with a negative slope equal to $-k$.

(A) For a first order reaction,the relationship between the rate constant $k$ and the half-life $t_{1/2}$ is given by the formula:
$k = \frac{0.693}{t_{1/2}}$
Given $t_{1/2} = 1200 \ s$.
Substituting the value:
$k = \frac{0.693}{1200} \ s^{-1}$
$k = 0.0005775 \ s^{-1}$
Rounding to two significant figures,we get:
$k \approx 5.8 \times 10^{-4} \ s^{-1}$

(C) For a first order reaction,the integrated rate equation is:
$\ln[A]_t = -kt + \ln[A]_0$
Converting to base $10$ logarithm:
$\log[A]_t = \frac{-kt}{2.303} + \log[A]_0$
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log[A]_t$,$x = t$,and $c = \log[A]_0$,the slope $m$ is equal to $\frac{-k}{2.303}$.

(B) For a first order reaction,the relationship between half-life $(t_{1/2})$ and rate constant $(k)$ is given by $t_{1/2} = \frac{0.693}{k}$.
For reaction $(1)$: $A \rightarrow B$,$k = 0.693 \ min^{-1}$. Therefore,$t_{1/2} = \frac{0.693}{0.693} = 1.0 \ min$.
For reaction $(2)$: $A \rightarrow C$,$t_{1/2} = 0.693 \ min$. Therefore,$k = \frac{0.693}{0.693} = 1.0 \ min^{-1}$.
Comparing the rate constants,for reaction $(1)$,$k = 0.693 \ min^{-1}$ and for reaction $(2)$,$k = 1.0 \ min^{-1}$.
Since the rate of a first order reaction is $Rate = k[A]$,and the initial concentration of $A$ is the same for both,the reaction with the higher rate constant is faster.
Since $1.0 > 0.693$,reaction $(2)$ is faster than reaction $(1)$,meaning reaction $(1)$ is slower than reaction $(2)$.

(A) For a first order reaction,the rate equation is $t = \frac{2.303}{k} \log \frac{100}{100 - x}$.
For $10 \%$ completion $(x = 10)$: $20 = \frac{2.303}{k} \log \frac{100}{90} = \frac{2.303}{k} \log (10/9)$ $(i)$.
For $19 \%$ completion $(x = 19)$: $t = \frac{2.303}{k} \log \frac{100}{100 - 19} = \frac{2.303}{k} \log \frac{100}{81} = \frac{2.303}{k} \log (10/9)^2 = 2 \times \frac{2.303}{k} \log (10/9)$ (ii).
Dividing equation (ii) by equation $(i)$:
$\frac{t}{20} = \frac{2 \times \frac{2.303}{k} \log (10/9)}{\frac{2.303}{k} \log (10/9)} = 2$.
Therefore,$t = 20 \times 2 = 40 \ min$.

(A) For a first-order reaction,the time $t$ taken for the concentration to reach $A_t$ from $A_0$ is given by $t = \frac{1}{k} \ln \frac{A_0}{A_t}$.
For $t_{1/8}$,the remaining concentration is $\frac{A_0}{8}$,so $t_{1/8} = \frac{1}{k} \ln \frac{A_0}{A_0/8} = \frac{1}{k} \ln 8$.
For $t_{1/10}$,the remaining concentration is $\frac{A_0}{10}$,so $t_{1/10} = \frac{1}{k} \ln \frac{A_0}{A_0/10} = \frac{1}{k} \ln 10$.
Taking the ratio: $\frac{t_{1/8}}{t_{1/10}} = \frac{\ln 8}{\ln 10} = \frac{\log 8}{\log 10} = \log 8$.
Since $\log 8 = \log 2^3 = 3 \log 2 = 3 \times 0.3 = 0.9$.
Therefore,$\frac{t_{1/8}}{t_{1/10}} \times 10 = 0.9 \times 10 = 9$.

(B) The reaction is $A_{(g)} \rightarrow B_{(g)} + C_{(g)}$.
Initially $(t=0)$: $P_A = 1 \ bar$,$P_B = 0$,$P_C = 0$. Total pressure $P_0 = 1 \ bar$.
At time $t = 100 \ min$: $P_A = 1 - x$,$P_B = x$,$P_C = x$.
Total pressure $P_t = (1 - x) + x + x = 1 + x$.
Given $P_t = 1.5 \ bar$,so $1 + x = 1.5 \implies x = 0.5 \ bar$.
Partial pressure of $A$ at $t = 100 \ min$ is $P_A = 1 - 0.5 = 0.5 \ bar$.
For a first-order reaction,$k = \frac{2.303}{t} \log \frac{P_0}{P_A}$.
$k = \frac{2.303}{100} \log \frac{1}{0.5} = \frac{2.303}{100} \log 2$.
Using $\log 2 = 0.3$,$k = \frac{2.303 \times 0.3}{100} = \frac{0.6909}{100} \approx 6.9 \times 10^{-3} \ min^{-1}$.

(D) For a first-order reaction,the rate constant is given as $k = 2.0 \text{ min}^{-1}$.
The formula for the half-life $(t_{1/2})$ of a first-order reaction is $t_{1/2} = \frac{0.693}{k}$.
Substituting the value of $k$: $t_{1/2} = \frac{0.693}{2.0} = 0.3465 \text{ min}$.
To convert the half-life from minutes to seconds,multiply by $60$: $0.3465 \text{ min} \times 60 \text{ s/min} = 20.79 \text{ s}$.
Rounding to one decimal place,we get $20.8 \text{ seconds}$.

(A) $1$. Consider the reaction: $A(g) \to B(g) + C(g)$.
$2$. At $t = 0$,the initial pressure of $A$ is $p_i$,and the pressures of $B$ and $C$ are $0$.
$3$. At time $t$,let $x$ be the decrease in pressure of $A$. The pressures are: $P_A = p_i - x$,$P_B = x$,and $P_C = x$.
$4$. The total pressure $p_t$ at time $t$ is given by: $p_t = (p_i - x) + x + x = p_i + x$.
$5$. From this,we find $x = p_t - p_i$.
$6$. The partial pressure of $A$ at time $t$ is $P_A = p_i - x = p_i - (p_t - p_i) = 2p_i - p_t$.
$7$. For a first-order reaction,the rate constant $k$ is given by: $k = \frac{1}{t} \ln \frac{p_i}{P_A}$.
$8$. Substituting $P_A$,we get: $k = \frac{1}{t} \ln \frac{p_i}{2p_i - p_t}$.

(C) $1$. For a first order reaction,the concentration of reactant at time $t$ is given by $[R]_t = [R]_0 e^{-kt}$.
$2$. The fraction of reactant remaining at time $t$ is the ratio of the concentration at time $t$ to the initial concentration: $\frac{[R]_t}{[R]_0} = e^{-kt}$.
$3$. The fraction of molecules decomposed is equal to $1$ minus the fraction remaining: $\text{Fraction decomposed} = 1 - \frac{[R]_t}{[R]_0} = 1 - e^{-kt}$.

(A) For a first-order reaction,the amount remaining after $n$ half-lives is given by the formula $N = N_0 \times (1/2)^n$.
Here,the total time is $t = 6 \text{ hours}$ and the half-life is $t_{1/2} = 3 \text{ hours}$.
The number of half-lives $n$ is calculated as $n = t / t_{1/2} = 6 / 3 = 2$.
The fraction of sucrose remaining is $(1/2)^n = (1/2)^2 = 1/4 = 0.25$.
To find the percentage remaining,we multiply the fraction by $100$: $0.25 \times 100 = 25\%$.
Thus,$25\%$ of the sucrose remains after $6 \text{ hours}$.

(A) $1$. For a first-order reaction,the rate constant $k$ is given by $k = 0.693 / t_{1/2}$.
Given $t_{1/2} = 6.93$ minutes,so $k = 0.693 / 6.93 = 0.1 \text{ min}^{-1}$.
$2$. The time $t$ required for the completion of $99\%$ of the reaction is calculated using the formula: $t = (2.303 / k) \log([A]_0 / [A]_t)$.
Here,$[A]_0 = 100$ and $[A]_t = 100 - 99 = 1$.
$3$. Substituting the values: $t = (2.303 / 0.1) \log(100 / 1) = 23.03 \times \log(10^2) = 23.03 \times 2 = 46.06$ minutes.
Rounding to the nearest integer,the time is $46$ minutes.

(A) For a first-order reaction,the rate constant $k$ is given by the formula:
$k = \frac{1}{t} \ln\left(\frac{[A]_0}{[A]_t}\right)$
At $t = 20 \text{ min}$,$[A]_0 = 0.6500 \text{ M}$ and $[A]_t = 0.00065 \text{ M}$.
$k = \frac{1}{20} \ln\left(\frac{0.6500}{0.00065}\right) = \frac{1}{20} \ln(1000) = \frac{1}{20} \times 6.908 = 0.3454 \text{ min}^{-1}$.
Now,for $t = x$,$[A]_t = 0.0650 \text{ M}$.
$k = \frac{1}{x} \ln\left(\frac{[A]_0}{[A]_t}\right)$
$0.3454 = \frac{1}{x} \ln\left(\frac{0.6500}{0.0650}\right)$
$0.3454 = \frac{1}{x} \ln(10)$
$0.3454 = \frac{2.303}{x}$
$x = \frac{2.303}{0.3454} \approx 6.67 \text{ min}$.
Rounding to the nearest integer,we get $x = 7 \text{ min}$. However,looking at the options provided,there might be a discrepancy in the question data or options. If we consider the ratio of concentrations,$[A]_0/[A]_x = 10$ and $[A]_0/[A]_{20} = 1000 = 10^3$. Since the reaction is first order,the time taken for a concentration change of $10^n$ is proportional to $n$. Thus,if $20 \text{ min}$ corresponds to $n=3$,then $x$ (for $n=1$) should be $20/3 \approx 6.67 \text{ min}$. Given the options,$x = 7$ is the closest integer.