Chemical Kinetics

From the balanced reaction,

$ 2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3 $

Using stoichiometry,

$ \begin{aligned} & \text { Rate of disapperance of } \mathrm{O}_2=2 \times 10^{-1} \mathrm{molL}^{-1} \mathrm{~s}^{-1} \\ &\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\frac{-d\left[\mathrm{O}_2\right]}{d t}=\frac{1}{2} \frac{d\left[\mathrm{SO}_3\right]}{d t} \\ &\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\, \frac{d\left[\mathrm{SO}_3\right]}{d t}=2 \times\left[\frac{-d\left[\mathrm{O}_2\right]}{d t}\right] \end{aligned} $

Rate of appearance of

$ \mathrm{SO}_3=2 \times\left(2 \times 10^{-1}\right)=4 \times 10^{-1} \mathrm{molL}^{-1} \mathrm{~s}^{-1} $

Half-life for first-order reaction,

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

Given, $t_{1 / 2}=600 \mathrm{~s}$

$ k=\frac{0.693}{600} \approx 1.155 \times 10^{-3} \mathrm{~s}^{-1} $

Using Arrhenius,

$ k=A e^{\left(-E_0 / R T\right)} $

Take $\log$ both sides and solve,

$ \ln k=\ln (A)-\frac{E_a}{R T} $

$ \begin{aligned} &\text { Substituting the values, }\\ &\begin{aligned} \ln \left(1.155 \times 10^{-3}\right) & =\ln \left(4 \times 10^{13}\right)-\left(\frac{98600}{R T}\right) \\ & \approx 311.49 k \end{aligned} \end{aligned} $

For the reaction $ 2 \mathrm{SO}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3} $, we can determine the rate of appearance of $\mathrm{SO}_{3}$ from the given information.

Given:

Rate of disappearance of $\mathrm{O}_{2}$ is $2 \times 10^{-4} \, \mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{s}^{-1}$.

The reaction rate expressions can be set up as:

$ \text{Rate of reaction} = -\frac{1}{2} \frac{d[\mathrm{SO}_{2}]}{dt} = -\frac{d[\mathrm{O}_{2}]}{dt} = \frac{1}{2} \frac{d[\mathrm{SO}_{3}]}{dt} $

To find the rate of appearance of $\mathrm{SO}_{3}$, we use the stoichiometry of the reaction. Since one mole of $\mathrm{O}_{2}$ produces two moles of $\mathrm{SO}_{3}$, the rate of appearance of $\mathrm{SO}_{3}$ is twice the rate of disappearance of $\mathrm{O}_{2}$.

$ \frac{d[\mathrm{SO}_{3}]}{dt} = 2 \times \left(-\frac{d[\mathrm{O}_{2}]}{dt}\right) = 2 \times (2 \times 10^{-4}) $

Calculating this gives:

$ \frac{d[\mathrm{SO}_{3}]}{dt} = 4 \times 10^{-4} \, \mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{s}^{-1} $

Thus, the rate of appearance of $\mathrm{SO}_{3}$ is $4 \times 10^{-4} \, \mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{s}^{-1}$.

According to the Arrhenius equation:

$ \log k = \log A - \frac{E_a}{2303RT} $

For a first-order reaction, the rate constant $ k $ can be expressed in terms of half-life ($ t_{1/2} $) as:

$ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{600 \text{ s}} = 1.1 \times 10^{-3} \text{ s}^{-1} $

Next, we substitute the value of $ k $ into the Arrhenius equation:

$ \log \left(1.1 \times 10^{-3}\right) = \log \left(4 \times 10^{13}\right) - \frac{98.6 \times 10^3}{2.303 \times 8.314 \times T} $

Solving for $ T $, we find:

$ T = 311.15 \text{ K} $

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.

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