Chemical Kinetics

The change in concentration, $\Delta[R]$ is the final concentration minus initial concentration.

$ \Delta[R]=0.03-0.04=-0.01 \mathrm{~mol} / \mathrm{L} $

$ \begin{aligned} \text { Average velocity } & =-\frac{\Delta[R]}{\Delta t}=-\left(-\frac{0.01}{40}\right) \\ & =0.00025 \mathrm{~mol}^{-1} \mathrm{~min}^{-1} \end{aligned} $

or $2.5 \times 10^{-4} \mathrm{~mol}^{-1} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$

In $\mathrm{Mol}^{-1} \mathrm{~L}^{-1} \mathrm{~s}^{-1}=4.167 \times 10^{-6} \mathrm{~mol}^{-1} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$

The slope of the $\ln K$ vs $\frac{1}{T}$ plot is given by

$ \begin{aligned} \text { Slope }= & -\frac{E_a}{R} \Rightarrow-2 \times 10^4=\frac{E_a}{8.314} \\ E_a & =166280 \mathrm{~J} / \mathrm{mol} \\ & =166.28 \mathrm{~kJ} / \mathrm{mol} \approx 166 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $

From the graph,

Energy of reactant $A=10 \mathrm{~kJ} / \mathrm{mol}$

Energy of product $P=5 \mathrm{~kJ} / \mathrm{mol}$

Energy of transition state (peak) $=25 \mathrm{~kJ} / \mathrm{mol}$.

Activation energy $E_a=$ Energy of transition sate - Energy of reactant

Heat of reaction $\Delta H=$ Energy of product

- Energy of reactant

$\Delta H=5-10=-5 \mathrm{~kJ} / \mathrm{mol}$

$|\Delta H|=5 \mathrm{~kJ} / \mathrm{mol}$

For half completion $[A]_t=\frac{[A]_0}{2}$

$ t_{1 / 2}=\frac{2.303}{K} \log \frac{[A]_0}{[A]_{0 / 2}} $

Simplify $t_{1 / 2}=\frac{0.693}{K}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

for $3 / 4$ completion

$ \begin{aligned} {[A]_t } & =\frac{1}{4}[A]_0 \\ t_{3 / 4} & =2 \times t_{1 / 2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Taking ratio of Eqs. (i) and (ii),

$ \frac{t_{3 / 4}}{t_{1 / 2}}=\frac{2 \times t_{1 / 2}}{t_{1 / 2}}=2: 1 $

The given equation is

$ \log _{10} \frac{k}{A}=0.00174 $

Multiply 2.303 on both side

$ \begin{aligned} \quad \ln \left(\frac{k}{A}\right) & =2.303 \times 0.00714 \\ \Rightarrow \quad \ln \left(\frac{k}{A}\right) & =0.00400722\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

The Arrhenius equation is

$ k=A e^{-E_e / R T} \Rightarrow \frac{k}{A}=e^{-E a / R T} $

Taking natural log

$ \ln \left(\frac{k}{A}\right)=-\frac{E_a}{R T} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Compare Eqs. (i) and (ii)

$ -\frac{E_a}{R T}=0.00400722 $

Substituting all values

$ E_a \simeq 10.0 \mathrm{~J} / \mathrm{mol} $

We want to find the average velocity (rate) for the reaction as $A$ goes to $B$ when the amount of $A$ drops from $x$ to $y$ mol/L in 100 minutes.

The change in the concentration of $A$ is: $ \Delta [A] = y - x \text{ mol/L} $ This tells us how much $A$ has changed in value.

The average velocity means how much $A$ changes per minute. We can find it by dividing the total change in $A$ by the time taken: $ \text{Average velocity} = \frac{\Delta [A]}{\Delta t} $ where $\Delta t = 100$ minutes.

So, $ \text{Average velocity} = \frac{y - x}{100} \text{ mol L}^{-1} \text{ min}^{-1} $ If you need the answer as a positive value (since rates are usually positive), write it as $ \frac{|x - y|}{100} \text{ mol L}^{-1} \text{ min}^{-1} $

The concentration becomes $\frac{1}{8}$ of what it was at the start.

$ \frac{1}{8} = \left(\frac{1}{2}\right)^3 $

This means it takes 3 half-lives for the concentration to become $\frac{1}{8}$.

$ \begin{array}{r} t_{1 / 2} = \frac{75}{3} \\ t_{1 / 2} = 25\ \text{minutes} \end{array} $

Given equation,

$ \log \left(\frac{K}{A}\right)=-\frac{x}{T} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Using Arrhenius equation

$ k=A e^{-E_q / R T} $

Taking log on both side

$ \begin{aligned} & \log k=\log A-\frac{E_a}{2.303 R T} \\ & \log \left(\frac{k}{A}\right)=-\frac{E_a}{2.303 R T}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

Compare Eqs. (i) and (ii),

$ \frac{x}{T}=\frac{E_a}{2.303 R T} \Rightarrow E_a=2.303 R \cdot x $

Given reaction is,

$ A+2 B \rightarrow 3 C+2 D $

The rate of reaction is, $r=-\frac{1}{2} \frac{d[B]}{d t}$

Given, $\frac{d[B]}{d t}=-x \times 10^{-2}$

Thus, $r=\frac{1}{2} \times x \times 10^{-2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Rate of appearance of $C$

$ \begin{aligned} & =\frac{1}{3} \frac{d[C]}{d t}=-\frac{1}{2} \frac{[d B]}{d t} \\ \frac{d[C]}{d t} & =\frac{3}{2} \times x \times 10^{-2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Taking the ratio of (i) and (ii)

$ \frac{\text { rate of reaction }}{\text { rate of appearance of } C}=1: 3 $

The activation energy for acid hydrolysis of sucrose is $6.22 \mathrm{~kJ} / \mathrm{mol}$. The activation energy for sucrose by sucrase is $2.15 \mathrm{~kJ} / \mathrm{mol}$.

For 1st order rate is, Rate $=k[A]$

$ \frac{\text { Rate }_1}{\text { Rate }_2}=\frac{\left[A_1\right]}{\left[A_2\right]} $

For first two data points,

$ \begin{aligned} & & \frac{0.2}{0.4} & =\frac{0.02}{x} \\ \Rightarrow & & x & =0.04 \mathrm{~m} \end{aligned} $

For second and third data point,

$ \begin{aligned} & \frac{0.4}{1}=\frac{0.04}{y} \Rightarrow y=0.1 \mathrm{M} \\ & x: y \Rightarrow 0.04: 0.1 \Rightarrow 2: 5 \end{aligned} $

The instaneous lst order reaction rate is given as

$ \text { rate }=\frac{-d[A]}{d t} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

On $X$-axis $=$ time, $Y$-axis $=$ concentration

From graph slope $=-m$

i.e., concentration is decreasing with time.

$ -m=\frac{d[A]}{d t} \text { from graph }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) $

Equating (i) and (ii)

Rate of reaction $=m$

To find the approximate activation energy ($E_a$) for a first-order reaction where the rate constant doubles when the temperature changes from 300 K to 310 K, we can use the following relationship:

$ \log \left(\frac{K_2}{K_1}\right) = \frac{E_a}{2.303 R} \frac{T_2 - T_1}{T_2 \times T_1} $

For $T_1 = 300 \, \text{K}$, the rate constant is $K_1 = K$.

For $T_2 = 310 \, \text{K}$, the rate constant becomes $K_2 = 2K$.

By substituting the known values into the equation:

$ \log \left(\frac{2K}{K}\right) = \frac{E_a}{2.303 \times 8.314} \times \frac{310 - 300}{310 \times 300} $

Simplifying and solving the equation:

$ \log 2 = \frac{E_a}{2.303 \times 8.3} \times \frac{10}{310 \times 300} $

Given that $\log 2 = 0.3$, we rearrange to find $E_a$:

$ E_a = 5333 \, \text{J/mol} = 53.33 \, \text{kJ/mol} $

Thus, the approximate activation energy is $53.33 \, \text{kJ/mol}$.

For a first-order reaction, the integrated rate law is:

$ k = \frac{1}{t} \ln \frac{[A]_0}{[A]_t} \quad \text{...(i)} $

Where:

$ k $ is the rate constant,

$ [A]_0 $ is the initial concentration,

$ [A]_t $ is the concentration at time $ t $.

Given:

$ k = 3.3 \times 10^{-4} \, \mathrm{s}^{-1} $,

$ [A]_0 = 100 $ (assuming initial concentration as 100 for calculation),

$ [A]_t = 10 $ (since 90% of the reaction is complete, only 10% remains).

Substitute these values into equation (i):

$ t = \frac{1}{k} \ln \frac{[A]_0}{[A]_t} = \frac{1}{3.3 \times 10^{-4}} \ln \left(\frac{100}{10}\right) $

This simplifies to:

$ t = \frac{10^4 \times 2.303}{3.3} = 6977.53 \, \mathrm{s} $

Converting seconds to minutes:

$ \frac{6977.5}{60} = 116.67 \, \mathrm{min} $

So, the time required to complete 90% of the reaction is 116.67 minutes.

For a zero-order reaction, the rate of reaction is independent of the concentration of the reactant. Given the rate constant $ k = 1 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} $, the initial concentration of $ A $ is $ [A]_0 = 0.1 \, \text{mol L}^{-1} $, and the time $ t = 10 $ seconds, we use the following formula to calculate the concentration of $ A $ after 10 seconds:

$ k = \frac{[A]_0 - [A]_t}{t} $

Substituting the known values into the formula, we get:

$ 1 \times 10^{-3} = \frac{0.1 - [A]_t}{10} $

Solving for $ [A]_t $:

$ 0.01 = 0.1 - [A]_t $

$ [A]_t = 0.1 - 0.01 $

$ [A]_t = 0.09 \, \text{mol L}^{-1} $

Therefore, the concentration of $ A $ after 10 seconds is $ 0.09 \, \text{mol L}^{-1} $.

To determine the rate constant of a reaction at a different temperature using the Arrhenius equation, follow this process:

The formula for calculating the change in the rate constant with temperature is:

$ \log \frac{K_{T_2}}{K_{T_1}} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) $

Where:

$ K_{T_1} $ is the rate constant at temperature $ T_1 $.

$ K_{T_2} $ is the rate constant at temperature $ T_2 $.

$ E_a $ is the activation energy.

$ R $ is the universal gas constant.

Given Values:

Initial rate constant, $ K_{T_1} = 3.46 \times 10^{-2} \, \mathrm{s}^{-1} $ at $ T_1 = 298 \, \mathrm{K} $

Activation energy, $ E_a = 50.1 \, \mathrm{kJ/mol} = 50.1 \times 10^3 \, \mathrm{J/mol} $

Universal gas constant, $ R = 8.314 \, \mathrm{J/K \cdot mol} $

Temperature, $ T_2 = 350 \, \mathrm{K} $

Calculation Steps:

Substitute the given values into the formula:

$ \log \frac{K_{T_2}}{3.46 \times 10^{-2}} = \frac{50.1 \times 10^3}{2.303 \times 8.314} \times \left( \frac{350 - 298}{298 \times 350} \right) $

Evaluate the expression:

$ \log \frac{K_{T_2}}{3.46 \times 10^{-2}} = 1.30 $

Solve for $ K_{T_2} $:

$ \frac{K_{T_2}}{3.46 \times 10^{-2}} = 10^{1.30} = 19.95 $

Calculate $ K_{T_2} $:

$ K_{T_2} = 3.46 \times 10^{-2} \times 19.95 = 0.6903 \, \mathrm{s}^{-1} $

Thus, the rate constant at 350 K is approximately $ 0.692 \, \mathrm{s}^{-1} $.

For the given reaction:

$ 5 \mathrm{Br}^{-}(aq) + \mathrm{BrO}_3^{-}(aq) + 6 \mathrm{H}^{+}(aq) \longrightarrow 3 \mathrm{Br}_2(aq) + 3 \mathrm{H}_2 \mathrm{O}(l) $

The rate of change for the concentration of $\mathrm{Br}^{-}$ ions is given as:

$ -\frac{\Delta [\mathrm{Br}^{-}]}{\Delta t} = x \, \text{mol L}^{-1} \text{min}^{-1} $

The overall rate of the reaction can be calculated by considering the stoichiometry of $\mathrm{Br}^{-}$. Since 5 moles of $\mathrm{Br}^{-}$ are consumed per mole of reaction, the rate of the reaction is:

$ \frac{\Delta R}{\Delta t} = -\frac{1}{5} \frac{\Delta [\mathrm{Br}^{-}]}{\Delta t} $

$ = -\frac{1}{5} \times (-x) = \frac{x}{5} \, \text{mol L}^{-1} \text{min}^{-1} $

Therefore, the rate of the reaction is $\frac{x}{5} \, \text{mol L}^{-1} \text{min}^{-1}$.

The average rate of a first-order reaction is calculated using the formula:

$ \frac{\Delta R}{\Delta t} = -\left[\frac{R_2 - R_1}{t_2 - t_1}\right] $

In this context:

$ R_1 $ is the initial concentration:

$ R_1 = 0.03 \, \text{mol/L} $

$ R_2 $ is the final concentration:

$ R_2 = 0.02 \, \text{mol/L} $

$ t_2 - t_1 $ is the time interval over which the change occurs, given as 25 minutes. To convert this time into seconds:

$ t_2 - t_1 = 25 \times 60 \, \text{seconds} $

By substituting these values into the rate formula, we obtain:

$ \frac{\Delta R}{\Delta t} = -\frac{0.02 - 0.03}{25 \times 60} $

Calculating the above expression provides:

$ \frac{\Delta R}{\Delta t} = 6.667 \times 10^{-6} \, \text{mol/L/s} $

For a first-order reaction,

$ A(g) \rightarrow B(g) + 2C(g) $

After 24 minutes, the ratio of the concentration of the products to the reactant is $1:3$.

The rate constant $k$ for a first-order reaction is given by the formula:

$ k = \frac{2.303}{t} \log \left[\frac{a}{a-x}\right] $

Where:

$ t = 24 $ minutes

$ a = 4 $ (initial concentration)

$ a-x = 3 $

Substituting these into the equation, we have:

$ k = \frac{2.303}{24} \log 1.11 $

Using $\log 1.11 = 0.046$, we get:

$ k = \frac{2.303}{24} \times 0.046 \approx 4.40 \times 10^{-3} \, \text{min}^{-1} $

The half-life ($t_{1/2}$) for a first-order reaction is calculated using:

$ t_{1/2} = \frac{0.693}{k} $

Substitute $k = 4.40 \times 10^{-3}$ min$^{-1}$:

$ t_{1/2} = \frac{0.693}{4.40 \times 10^{-3}} \approx 157.8 \, \text{minutes} $

A reaction is described as autocatalytic when one of its products also serves as a catalyst for the same reaction.

For the given reaction:

$ \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5 + \mathrm{H}_2 \mathrm{O} \xrightarrow{\mathrm{H}^{+}} \mathrm{CH}_3 \mathrm{COOH} + \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} $

Here, $\mathrm{CH}_3 \mathrm{COOH}$ is both a product and a catalyst for the reaction, illustrating the autocatalytic nature.

Given, first order reaction graph between $\ln (a-x)$ on $Y$-axis and $t$ on $X$-axis.

According to first order reaction, $\ln (a-x)$

Rate constant is given by

$ k=\frac{\ln }{t} \frac{[a]}{[a-x]} \quad \ldots \text { (i) }$

or $ \begin{aligned} k t & =\ln \frac{[a]}{[a-x]} \\ k t & =\ln [a]-\ln [a-x] \end{aligned} $

or $\ln [a-x]=\ln [a]-k t \quad \ldots \text { (ii) } $

Compare Eq. (ii) by straight line equation.

$ \begin{aligned} & Y=\ln [a-x], x=t, m(\text { slope })=-k \\ & C \text { (intercent })=\ln \text { al } \end{aligned} $

$C$ (intercept) $=\ln [a]$

Therefore, slope $=$ rate constant

$ =-\left(-10^{-2}\right)=10^{-2} \quad \ldots \text { (iii) } $

Now initial concentration (time $t=0$ ) Eq. (ii) becomes

$ \ln [a-x]=\ln [a] $

From graph at $t=0, \ln [a-x]=-2.303$

$ -2.303=\ln [a] $

Initial concentration, $[a]=10^{-1}$

Given,

$\begin{aligned} & k_1=0.02 \mathrm{~s}^{-1} \\ & k_2=0.2 \mathrm{~s}^{-1} \\ & E_d=? \\ & R=8.3 \\ & T_1=500 \mathrm{~K} \\ & T_2=700 \mathrm{~K} \end{aligned}$

$\begin{aligned} & \text { In } \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{r_2}-\frac{1}{T_1}\right) \\ & \text { In } \frac{0.2}{0.02}=-\frac{E_a}{8.3}\left[\frac{1}{700}-\frac{1}{500}\right] \\ & \text { In } 10=\frac{-E_a}{8.3}\left(\frac{500-700}{700 \times 500}\right) \\ & 2.302 \times 8.3 \times 700 \times 500=200 E_a \\ & \Rightarrow E_a=\frac{2.302 \times 8.3 \times 700 \times 500}{200} \\ & =33436 \mathrm{~J} / \mathrm{mol} \\ & \approx 33.44 \mathrm{~kJ} / \mathrm{mol} \\ \end{aligned}$

Conc. at time, $t, A=A_0-\frac{93.75}{100} A_0=\frac{6.25}{100} A_0$

$A_0 \rightarrow$ initial conc. $k \rightarrow$ rate constant, time, $t=x \mathrm{~min}$ Using integrated rate law for lst order reaction,

$\begin{aligned} k & =\frac{2.303}{t} \log \frac{A_0}{A} \\ & =\frac{2.303}{x} \log \frac{100 A_0}{6.25 A_0} \end{aligned}$

$\text { So, } \quad k=\frac{2.303}{x} \log \frac{100}{6.25}$

$\begin{aligned} & =\frac{2.303 \times 1.20}{x} \\ \Rightarrow \quad k & =\frac{2.303 \times 1.20}{x} \min ^{-1} \end{aligned}$

We know,

$\begin{aligned} t_{1 / 2} & =\frac{0.693}{k} \\ & =\frac{0.693 x}{2.303 \times 1.20} \end{aligned}$

So, $\quad t_{1 / 2}=\frac{x}{4} \min$

Given, rate constant, $k_0=0.003 \mathrm{molL}^{-1} \mathrm{~s}^{-1}$

Initial conc., $A_0=0.1 \mathrm{M}$

Final conc., $A=0.075 \mathrm{M}$

Using integrated rate law for zeroth order,

$\begin{aligned} A_0-A & =k t \\ 0.1-0.075 & =0.003 \times t \\ t & =\frac{0.025}{0.003}=8.33 \mathrm{~s} \end{aligned}$

Given:

$ \text{Rate} = k [A][B]^2 $

Step 1: Write dimensions of each term

• Rate of reaction has units:

  $ \text{Rate} = \mathrm{mol\,L^{-1}\,s^{-1}} $

• $[A]$ and $[B]$ (concentrations) both have units:

  $ \mathrm{mol\,L^{-1}} $

Step 2: Substitute into the rate equation

$ \mathrm{mol\,L^{-1}\,s^{-1}} = k \times (\mathrm{mol\,L^{-1}}) \times (\mathrm{mol\,L^{-1}})^2 $

Simplify right-hand side:

$ k \times \mathrm{mol^3\,L^{-3}} $

Step 3: Solve for units of $ k $

$ k = \frac{\mathrm{mol\,L^{-1}\,s^{-1}}}{\mathrm{mol^3\,L^{-3}}} $

$ k = \mathrm{mol^{-2}\,L^{2}\,s^{-1}} $

✅ Final Answer:

$ \boxed{k = \mathrm{mol^{-2}\,L^{2}\,s^{-1}}} $

Correct option: D

$\begin{aligned} & X(g) \longrightarrow Y(g)+Z(g) \\ & t_{1 / 2}=10 \mathrm{~min} \quad \text{(Given)}\\ & A=A_0 e^{-k t} \quad \text{....... (i)}\\ & {\left[A_0\right.}=\text { Initial concentration }] \\ & k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{10}=0.0693 \end{aligned}$

From eq. (i),

$\begin{aligned} 0.1 A_0 & =A_0 e^{-0.0693 \times t} \\ \ln 0.1 & =-0.0693 \times t \\ -2.303 & =-0.0693 t \Rightarrow t=33.23 \mathrm{~min} . \end{aligned}$

(b) Unit of rate of reaction $\frac{d[A]}{d t}=k[A]$

$\text { Unit }=\mathrm{ms}^{-1}$

(a) Rate of disappearance

$\begin{aligned} & A \rightarrow B \\ & -\frac{d[A]}{d t} \Rightarrow \mathrm{ms}^{-1} \end{aligned}$

(d) For 1st order

$\begin{aligned} \frac{-d[A]}{d t} & =k[A] \\ \mathrm{ms}^{-1} & =k \mathrm{~m} \\ k & =\mathrm{s}^{-1} \end{aligned}$

For 2nd order

$\begin{aligned} \frac{-d[A]}{d t} & =k[A]^2 \\ \mathrm{~ms}^{-1} & =k[\mathrm{~m}]^2 \\ k & =\mathrm{m}^{-1} \mathrm{~s}^{-1} \end{aligned}$

So, $k$ depends on order of reaction and unit of $k$ for 1st order reaction is $\mathrm{s}^{-1}$.

For zero order reaction,

$[A]=[A_0]-kt$ ..... (i)

$[A]=\frac{[A_0]}{2}$ (half-life) ..... (ii)

From eqs. (i) and (ii)

$\frac{\left[A_0\right]}{2}=\left[A_0\right]-k t_{1 / 2} \Rightarrow t_{1 / 2}=\frac{[A]_0}{2 k}$

Plot of $t_{1 / 2}$ versus $\left[A_0\right]$ will be straight line.

Initial weight $\left(A_0\right)$ or $a$ is $4 \mathrm{~g}$

Final weight at time $t$ or after reduce $(a-x)$ or $0.2 \mathrm{~g}$

First order reaction; $t=\frac{2.303}{k} \log \left(\frac{a}{a-x}\right)$

$\begin{aligned} t & =\frac{2.303}{2303 \times 10^{-3}} \log \left(\frac{49}{0.29}\right) \\ t & =10^3 \times \log 20 \\ t & =1301 \mathrm{~s} \text { or } t=0.36 \mathrm{~h} \end{aligned}$