Chemical Kinetics

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

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

The rate of a chemical reaction can be expressed in terms of the rate of change of concentration of any reactant or product. According to the given reaction:

$ 2A + 3B \to 3C + 4D $

The rate of this reaction can be expressed as the rate of disappearance of the reactants A or B, or as the rate of appearance of the products C or D. Since A is being consumed, its concentration will decrease over time and because B, C, and D are all involved in the reaction in different stoichiometric ratios, their rates of consumption or production will be proportional to their stoichiometric coefficients.

The general relation between the rate of the reaction and the rate of change of concentration of any species is given by:

$ \text{Rate} = -\frac{1}{\text{stoichiometric coefficient}} \frac{d[\text{reactant}]}{dt} = \frac{1}{\text{stoichiometric coefficient}} \frac{d[\text{product}]}{dt} $

For reactant A with a stoichiometric coefficient of 2:

$ \text{Rate} = -\frac{1}{2} \frac{d[A]}{dt} $

For reactant B with a stoichiometric coefficient of 3:

$ \text{Rate} = -\frac{1}{3} \frac{d[B]}{dt} $

For product C with a stoichiometric coefficient of 3:

$ \text{Rate} = \frac{1}{3} \frac{d[C]}{dt} $

For product D with a stoichiometric coefficient of 4:

$ \text{Rate} = \frac{1}{4} \frac{d[D]}{dt} $

So the correct options that match the expression for the rate of reaction based on the appearance of products C and D are:

Option C: $ {1 \over 3}{{d[C]} \over {dt}} $

Option D: $ {1 \over 4}{{d[D]} \over {dt}} $

Note that when expressing the rate with respect to the reactants (A and B), the rate must include a negative sign because their concentrations decrease as the reaction proceeds. For products (C and D), the rate is positive because their concentrations increase.

∵ Rate $(r)=$ rate constant $(k)$ [conc.] ${ }^{\text {order }}$

$ k=\frac{\text { rate }}{[\text { conc. }]^{\text {order }(=n)}}=\frac{\mathrm{mol} \mathrm{~L}^{-1}}{\left[\mathrm{~mol} \mathrm{~L}^{-1}\right]^n} $

where, $n=$ order of reaction

$ \therefore \quad k=\mathrm{mol}^{1-n}, \mathrm{~L}^{n-1} \mathrm{~s}^{-1} $

For zero order ( $n=0$ )

$ \therefore \quad k=\mathrm{mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} $

Hence, option (a) is the correct answer.

$ \text { Given, In } k=5.0-\frac{12000}{T} $

$ \begin{aligned} &\text { Also, }\\ &\begin{aligned} \because & & \operatorname{In} k & =\log A-\frac{E_a}{R T} \\ & \text { and } & k & =A e^{-E_a / R T} \\ & \text { o, } & \log _e K & =\log _e A=\frac{E_a}{R T} \end{aligned} \end{aligned} $

where, $A=$ pre-exponential constant i.e. frequency factor

$ \begin{aligned} E_a & =\text { activation energy } \\ R & =\text { gas constant }=2 \mathrm{kcal} \mathrm{~mol}^{-1} \\ T_k & =\text { temperature } \\ k & =\text { rate constant } \end{aligned} $

Thus,

$ \begin{array}{rlrl} & \frac{E_a}{R T} =\frac{12000}{T} \\ \text { or } & E_a =12000 \times R \\ \therefore & E_a =\frac{12000 \times 2}{1000}\,\,\,\left(\text { in } \mathrm{kcal} \mathrm{~mol}^{-1}\right) \\ \text { i.e. } & E_a =24 \mathrm{kcal} \mathrm{~mol}^{-1} \end{array} $

Hence, option (a) is the correct answer.

Any reaction of zero order must obey the following equation

$ \begin{aligned} {[A] } & =-k t+[A]_0 \\ \text { or }[A] & =[A]_0+k t \end{aligned} $

This shows that, the concentration of reactant decreases linearly with time, as it is an equation of a straight line $(y=m x+c)$, the plot of $[A]$ versus $t$ will be straight line with slope $=-k$ and intercept on the concentration axis $=[A]_0$ as shown in figure.

Plot of $[A]$ versus $t$ for a reaction of zero order.

From molarity equation,

M₁V₁ = M₂V₂

M₁ $\times$ 50.0 = 0.10 $\times$ 39.30

$\therefore$ ${M_1} = {{0.10 \times 39.30} \over {50}} = 0.0786$ M

When half equivalence point is reached,

$[HA] = [{A^ - }]$

$pH = p{K_a} + \log {{[{A^ - }]} \over {[HA]}}$

$pH = p{K_a} = 4.85$

P_t = 0.4 $-$ x + x + x + x = 0.4 + 2x or x = ${{{P_t} - 0.4} \over 2}$

P_A = P₀ $-$ x = P₀ $-$ ${{{P_t} - 0.4} \over 2}$ = ${{2 \times 0.4 - {P_t} + 0.4} \over 2} = {{1.2 - {P_t}} \over 2}$

From, $k = {{2.303} \over t}\log {{{P_0}} \over {{P_A}}}$

${{0.693} \over {{t_{1/2}}}} = {{2.303} \over {230}}\log \left( {{{0.4 \times 2} \over {1.2 - {P_t}}}} \right)$

${{0.693} \over {{t_{1/2}}}} = {{2.303} \over {230}}\log \left( {{{0.8} \over {1.2 - {P_t}}}} \right)$

$\log {{0.8} \over {1.2 - {P_t}}} = {{0.693 \times 230} \over {69.3 \times 2.303}} = 0.9987$

${{0.8} \over {1.2 - {P_t}}} = {10^{0.9987}}$

$0.8 = (10)(1.2 - {P_t})$

$0.8 = 12 - 10{P_t}$

$10{P_t} = 12 - 0.8$

$10{P_t} = 11.2$

${P_t} = {{11.2} \over {10}} = 1.12$ atm