Chemical Kinetics

Chemistry I

Chemistry

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How does temperature affect the rate of a reaction? Is there a corresponding equal decrease in number of collisions among molecules of a gaseous reaction? How is this effect explained by the concept of activation energy?

Temperature usually has a major effect on the rate of a chemical reaction. Molecules at a higher temperature have more thermal energy. Although collision frequency is greater at higher temperatures. When the temperature of reaction mixture is increased, the kinetic energy of molecules increases which results in increase in velocity of molecules which in turn, results in increase in number effective, collisions; For every 10°C rise in temperature the rate of the reaction becomes double-fold to five-fold. There is no corresponding increase in the number of collision among the gaseous molecules because for 10°C rise in temperature the increase in number of collisions is only 2 to 3 percent. This many fold increase in the rate of the reaction is explained by the concept of activation energy. Only those collisions are effective and result in the formation of product where the molecules possess a certain minimum amount of energy over and above their average energy which is called the activation energy. For every 10°C rise in temperature, the number of activated molecules increase by 200 to 500% and therefore the reaction rate becomes double to five fold.

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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

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

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

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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

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

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 âˆ†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