First Order reaction Questions in English

(D) For a first-order reaction, the rate constant $k$ is given by: $k = \frac{2.303}{t} \log_{10} \left( \frac{[A]_0}{[A]_t} \right)$.
Given: $[A]_t = 20 \% \text{ of } [A]_0 = 0.2 [A]_0$, so $\frac{[A]_0}{[A]_t} = \frac{1}{0.2} = 5$.
Given $t = 10 \text{ min}$ and $\log_{10}(5) = 0.6989$.
Substituting the values: $k = \frac{2.303}{10} \times 0.6989$.
$k = 0.2303 \times 0.6989 \approx 0.1609 \text{ min}^{-1}$.

(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 $t_{1/2} = 20 \text{ min}$.
Substituting the value into the formula:
$k = \frac{0.693}{20 \text{ min}}$
$k = 0.03465 \text{ min}^{-1}$
Rounding to four decimal places, we get $k \approx 0.0347 \text{ min}^{-1}$.

(D) For a first-order reaction, the half-life $t_{1/2}$ is given as $16 \text{ minutes}$.
In $32 \text{ minutes}$, the number of half-lives elapsed is $n = \frac{32}{16} = 2$.
The fraction of reactant remaining after $n$ half-lives is given by $(\frac{1}{2})^n$.
Remaining fraction $= (\frac{1}{2})^2 = \frac{1}{4}$.
The fraction that has reacted is $1 - \text{remaining fraction} = 1 - \frac{1}{4} = \frac{3}{4}$.

(B) Step $1$: Identify the unit of the rate constant $k$. The given unit is $\text{s}^{-1}$.
Step $2$: Recall the general formula for the units of the rate constant for a reaction of order $n$: $\text{unit} = (\text{concentration})^{1-n} \times \text{time}^{-1}$.
Step $3$: For a first-order reaction $(n=1)$, the unit is $(\text{mol L}^{-1})^{1-1} \times \text{s}^{-1} = \text{s}^{-1}$.
Step $4$: Since the given unit $\text{s}^{-1}$ matches the unit of a first-order reaction, the reaction is of the first order.

(A) For a first-order reaction, the rate law is given by: $\text{Rate} = k[A]$.
Given: $\text{Rate} = 5.4 \times 10^{-6} \text{ mol dm}^{-3} \text{ s}^{-1}$ and $[A] = 0.3 \text{ M} = 0.3 \text{ mol dm}^{-3}$.
Rearranging the formula for the rate constant $k$: $k = \frac{\text{Rate}}{[A]}$.
Substituting the values: $k = \frac{5.4 \times 10^{-6}}{0.3} \text{ s}^{-1}$.
$k = 18 \times 10^{-6} \text{ s}^{-1} = 1.8 \times 10^{-5} \text{ s}^{-1}$.

(A) For a $1^{st}$ order reaction, the integrated rate equation is $[R] = [R]_0 e^{-kt}$.
$(1)$ The plot of $[R]$ vs $t$ is an exponential decay curve, not a straight line.
$(2)$ The plot of $\ln[R]$ vs $t$ is a straight line with slope $-k$ and intercept $\ln[R]_0$.
$(3)$ The plot of $\log_{10}[R]$ vs $t$ is a straight line with slope $-\frac{k}{2.303}$ and intercept $\log_{10}[R]_0$.
$(4)$ The plot of $\log_{10}\frac{[R]_0}{[R]}$ vs $t$ is a straight line with slope $\frac{k}{2.303}$ passing through the origin.
Since the graph in option $A$ is shown as a horizontal straight line (which represents a zero-order reaction), it is not applicable for a $1^{st}$ order reaction.

(B) For a $1^{st}$ order reaction, the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t}$.
Substituting the values: $k = \frac{2.303}{40} \log \frac{0.1}{0.025} = \frac{2.303}{40} \log 4 = \frac{2.303 \times 0.6021}{40} \approx 0.03466 \text{ min}^{-1}$.
The rate of reaction is given by $\text{Rate} = k[R]$.
For $[R] = 0.01 \text{ M}$, $\text{Rate} = 0.03466 \times 0.01 = 3.466 \times 10^{-4} \text{ M min}^{-1} \approx 3.47 \times 10^{-4} \text{ M min}^{-1}$.