Chemical Kinetics

Chemistry I

Chemistry

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Consider a typical first order gas phase reaction

$\mathrm{A}\left(\mathrm{g}\right)\to \mathrm{B}\left(\mathrm{g}\right)+\mathrm{C}\left(\mathrm{g}\right)$

P_{1} be the initial pressure of A, P_{t} the total pressure at time ‘t’. Derive integrated rate equation.

$\mathrm{A}\left(\mathrm{g}\right)\to \mathrm{B}\left(\mathrm{g}\right)+\mathrm{C}\left(\mathrm{g}\right)$

P

Total pressure P_{1} = P_{A} + P_{B} ^{+} P_{c} (pressure units)

p_{A}, p_{B} and p_{c} are the partial pressures of A, B and C respectively. When x is the amount of A converted into products, when pi is the initial pressure at time t = 0

${p}_{t}=({p}_{i}-\mathrm{x})+\mathrm{x}+\mathrm{x}={p}_{i}+\mathrm{x}\phantom{\rule{0ex}{0ex}}\mathrm{x}={p}_{t}-{p}_{i}$

where

${\mathrm{p}}_{\mathrm{A}}={p}_{i}-\mathrm{x}={p}_{i}-({p}_{t}-{p}_{i})=2{p}_{i}-{p}_{t}\phantom{\rule{0ex}{0ex}}\mathrm{k}=\frac{2.303}{\mathrm{t}}\mathrm{log}\frac{{p}_{i}}{{p}_{A}}=\frac{2.303}{\mathrm{t}}\mathrm{log}\frac{{p}_{i}}{(2{p}_{i}-{p}_{t})}$

Pseudo first order reaction: Although in most reactions, order and molecularity are same, there are certain reactions whose order and molecularity differ. For example, hydrolysis of

sugarcane,

${\mathrm{C}}_{12}{\mathrm{H}}_{12}{\mathrm{O}}_{11}+{\mathrm{H}}_{2}\mathrm{O}\to {\mathrm{C}}_{6}{\mathrm{H}}_{12}{\mathrm{O}}_{6}+{\mathrm{C}}_{6}{\mathrm{H}}_{12}{\mathrm{O}}_{6}\phantom{\rule{0ex}{0ex}}\mathrm{Sucrose}\mathrm{Glucose}\mathrm{Fructose}$

Molecularity of this reaction is 2 but its order 1 because its rate depends only on the concentration of surcrose. The concentration of water remains is very high and does not change during the reaction (i.e., concentration of water remains practically constant throughout the reaction). Such reactions are known as pseudo-unimolecular or pseudo first order reactions. Other examples, of pseudo-unimolecular reaction is the acidic hydrolysis of esters where water

remains in excess.

${\mathrm{CH}}_{3}{\mathrm{COOC}}_{2}{\mathrm{H}}_{5}+{\mathrm{H}}_{2}\mathrm{O}\stackrel{\left(\mathrm{excess}\right){\mathrm{H}}^{+}}{\to}{\mathrm{CH}}_{3}\mathrm{COOH}+{\mathrm{C}}_{2}{\mathrm{H}}_{2}\mathrm{OH}$

Although it is termolecular (molecularity = 3) reaction, its order is one as concentration of H^{+} and H_{2}O^{+} remains constant during reaction. Hydrolysis of organic chlorides is also an example of first order reaction small water (one of the reactants) is again in large excess and its concentration remains constant throughout the reactions.

Thus when one of the reactants is present in large excess, the second order reaction conforms to the first order and is known as a pseudo-unimolecular reaction.

Reaction between acetic anhydride and excess of ethanol to form ester and conversion of N-Chloroacetanilide to p-chloroacetanilide are also examples of pseudo-unimolecular reactions.

286 Views

A first order reaction has a rate constant 1.15 x 10^{–3} s^{–1}. How long will 5 g of this reactant take to reduce to 3 g?

Given that

Initial quantity, [R]

Final quantity, [R] = 3 g

Rate constant, k = 1.15 x 10

Formula of 1

We know that

$\mathrm{t}=\frac{2.303}{\mathrm{k}}\mathrm{log}\frac{{\left[\mathrm{R}\right]}_{0}}{\left[\mathrm{R}\right]}$

or

$\mathrm{t}=\frac{2.303}{1.15\times {10}^{-3}}\mathrm{log}\left(\frac{5}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{2.303}{1.15\times {10}^{-3}}(\mathrm{log}5-\mathrm{log}3)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{2.303}{1.15\times {10}^{-3}}(0.6990-0.4771)\phantom{\rule{0ex}{0ex}}=\frac{2.303\times 0.2219}{1.15\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=\frac{2.303\times 0.2219\times 1000}{1.15}\phantom{\rule{0ex}{0ex}}=444\mathrm{sec}.$

$\mathrm{t}=\frac{2.303}{1.15\times {10}^{-3}}\mathrm{log}\left(\frac{5}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{2.303}{1.15\times {10}^{-3}}(\mathrm{log}5-\mathrm{log}3)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{2.303}{1.15\times {10}^{-3}}(0.6990-0.4771)\phantom{\rule{0ex}{0ex}}=\frac{2.303\times 0.2219}{1.15\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=\frac{2.303\times 0.2219\times 1000}{1.15}\phantom{\rule{0ex}{0ex}}=444\mathrm{sec}.$

1299 Views

The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

Let the reaction is X →Y

This reaction follows second order kinetics.

So that, the rate equation for this reaction will

Rate, R = k[X]^{2} .............(1)

Let initial concentration is x mol L^{−1},

Plug the value in equation (1)

Rate, R_{1} = k .(a)^{2}

= ka^{2}

Given that concentration is increasing by 3 times so new concentration will 3a mol L^{−1}

Plug the value in equation (1) we get

Rate, R_{2} = k (3a)^{2}

= 9ka^{2}

We have already get that R_{1} = ka_{2} plus this value we get

R_{2} = 9 R_{1}

So that, the rate of formation will increase by 9 times.

Rate = k[A]^{2}If concentration of X is increased to three times,

Rate = k[3A]^{2}or Rate = 9 k A^{2}Thus, rate will increase 9 times.

972 Views

In a reaction 2A → Products, the concentration of A decreases from 0.5 mol to 0.4 mol L^{–1} in 10 minutes. Calculate the rate during this interval.

Given that

Initial concentration [A

Final concentration [A

Time is = 10 min

Rate of reaction = Rate of disappearance of A.

Rate of reaction = $-\frac{1}{2}\frac{\u2206\left[\mathrm{A}\right]}{\u2206\mathrm{t}}\phantom{\rule{0ex}{0ex}}$

$=-\frac{1}{2}\frac{(0.4-0.5)\mathrm{mol}{\mathrm{L}}^{-1}}{10\mathrm{minute}}\phantom{\rule{0ex}{0ex}}=\frac{0.1\mathrm{mol}{\mathrm{L}}^{-1}}{5\mathrm{minutes}}\phantom{\rule{0ex}{0ex}}=0.005\mathrm{mol}{\mathrm{litre}}^{-1}{\mathrm{min}}^{-1}.$

2088 Views

For a reaction, A + B → Product; the rate law is given by, r = k[ A]^{1/2} [B]^{2}. What is the order of reaction?

The order of the reaction is sum of the powers on concentration.

So that sum will

r = k[A]

1504 Views

For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M is 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

Given that

Initial concentration, [R1] = 0.03

Final concentration, [R2] = 0.02

Time taken ∆t = 25 min = 25 × 6 0 = 1500 sec (1 min = 60 sec )

The formula of average rate of change

${\mathrm{r}}_{\mathrm{av}}=\frac{-\u2206\mathrm{R}}{\u2206\mathrm{t}}=\frac{\u2206\left[\mathrm{P}\right]}{\u2206\mathrm{t}}$

(i) Average rate

$=\frac{(0.03-0.02)\mathrm{M}}{25\times 60\mathrm{sec}}\phantom{\rule{0ex}{0ex}}=\frac{0.01\mathrm{M}}{25\times 60\mathrm{s}}=6.66\mathrm{M}{\mathrm{s}}^{-1}$

(ii) Average rate

$=\frac{(0.03-0.02)\hspace{0.17em}\mathrm{M}}{25\mathrm{min}}=\frac{0.01\mathrm{M}}{25}\phantom{\rule{0ex}{0ex}}=0.0004{\mathrm{Ms}}^{-1}.$

(i) Average rate

$=\frac{(0.03-0.02)\mathrm{M}}{25\times 60\mathrm{sec}}\phantom{\rule{0ex}{0ex}}=\frac{0.01\mathrm{M}}{25\times 60\mathrm{s}}=6.66\mathrm{M}{\mathrm{s}}^{-1}$

(ii) Average rate

$=\frac{(0.03-0.02)\hspace{0.17em}\mathrm{M}}{25\mathrm{min}}=\frac{0.01\mathrm{M}}{25}\phantom{\rule{0ex}{0ex}}=0.0004{\mathrm{Ms}}^{-1}.$

1717 Views