Question 1

Solution

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

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

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

Question 2

Solution

Given that

Initial concentration [A_{1}] =0.5

Final concentration [A_{2}] =0.4

Time is = 10 min

Rate of reaction = Rate of disappearance of A.

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

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

Initial concentration [A

Final concentration [A

Time is = 10 min

Rate of reaction = Rate of disappearance of A.

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

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

Question 3

Solution

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

So that sum will

r = k[A]^{1/2}[B]^{2}Order of reaction = $\frac{1}{2}+2=2.5.$^{}

So that sum will

r = k[A]

Question 4

Solution

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.