Chemical Kinetics

Chemistry I

Chemistry

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State the role of activated complex in a reaction and state its relation with activation energy?

Activated complex is the intermediate compound formed by reactants, which is highly unstable and readily changes into product. Those reactants which possess activation energy and collide in proper orientation can form activated complex which can easily form products.

Lower the activation energy, more easily activated complex will be formed and faster will be the reaction.

Activation energy = energy of activated complex - energy of reactants

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631 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 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

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

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