Chemical Kinetics

Chemistry I

Chemistry

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How can the rate of a fast reaction be determined by Flash Photolysis?

Flash photolysis: When we pass a powerful flash of short duration or laser beam, through the reaction mixture to initiate a reaction the atoms, ions or free radicals are formed.

These atoms, ions or free radicals formed can be identified by passing a second, Flash of light through the mixture immediately after the first flash.

For this the absorption spectrum of the mixture is monitored continuously at small regular intervals after first flash and the changes in the spectrum with time indicate the various processes occurring in the system.

It can also be studied by observing some other property like electrical conductance or magnetic property of the reaction mixture.

126 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

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

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

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