Class 12th

5.8

The precipitate which is obtained from a chemical reaction always contain some unwanted substances (eg ions, impurities) which get adsorbed onto the surface of the precipitate.

Therefore, it becomes important to wash the precipitate before estimating it quantitatively so as to remove these unwanted adsorbed substances and obtain accurate results.

4.28 Given:

Time t = 40 min

When 30% decomposition is undergone, 70% is the concentration.

We know, time taken  

t= 2.303/K log R₀ / R

Where, k- rate constant

[R]₀ - initial concentration

[R] - concentration at time 't'

40 = 2.303/K log R₀ / 0.7 R₀

40 = 2.303/K log 1 / 0.7

40 / 0.1549 = 2.303 / k

?258.23 = (2.303/k)

We know, Half-life t_1/2 = 0.693/k

Which can be written as, t_1/2 = 0.3010 * (2.303/k)

? t_1/2 = 0.3010 * 258.23

? t_1/2 = 77.72 min

4.25 Given:

Order of the reaction = 1

Let, Initial concentration [R]^°= x

Final concentration [R] = x/16

Rate constant k = 60 s^-1

We know, time

t= 2.303 / k log R₀ / R

t = 2.303 / 60 log (x/x/16)

t = 2.303 / 60s^-1 log (1/1/16)

t = 2.303 X log 16 / 60s^-1

Solving, we get t = 4.6 * 10^-2s

4.24

(iv) As log [N₂O₅] vs time is a straight line given reaction is first Hence its rate law will be, Rate = k [N₂O₅]

(v) The slope of above graph is slope = 0.000209 K = 303 * slope

⇒4.82 * 10^-4sec^-1

Now, t_1/2 = 0.693/K.

⇒0.693/4.82 * 10^-4

⇒t_1/2 = 1438 sec. which is almost equal to (ii)

4.23 Radio active decay occurs via first order rate law,

t_1/2 = 5730 years. rate constant (k) of given decay is 0.693/t_1/2

⇒0.693/5730 = 1.2 * 10^-4 year^-1

By first order integrated rate law age of the sample will be,

T  = ( 2.303 / 1.2 X 10^-4 ) log (A₀/A_t)

where T is the age of the sample, A₀ is the initial activity of the sample. and A_t is the activity of the sample at any time t

T  = ( 2.303 / 1.2 X 10^-4 ) log (A₀/0.8 A₀)

T = 0.18 * 10⁴ years.

Age of given sample is 0.18 * 10⁴ years.

4.22 Half life of first order reaction is, t_1/2 = ln2/K where t_1/2 is half life of first order reaction, K is rate constant of First order reaction.

(i) t_1/2 = ln2/200 s^-1

⇒t_1/2 = 0.693/200 s^-1 (? ln2 = 0.693)

⇒t_1/2 = 0.003465 sec.

(ii) t_1/2 = ln2/2 min^-1

⇒t_1/2 = 0.693/2 min^-1 (? ln2 = 0.693)

⇒t_1/2 = 0.3465 min

(iii) t_1/2 = ln2/4 year^-1

⇒t_1/2 = 0.693/4 years ^-1 (? ln2 = 0.693)

⇒t_1/2 = 0.17325 year.

Half life of 3 reactions are 0.003465 sec, 0.3465 min, 0.17325 year, respectively.

4.21 As reaction is first order with respect to A and zero Order with respect to B. Then changing the concentration of B won't affect the rate of reaction and increasing concentration of A 'n' times will increase the rate by 'n' times. By this logic lets fill the table- In first blank space concentration of A will be 0.2 mol L^-1 because the rate is doubled. In second blank space, Rate will be 8 * 10^-2mol L^-1min^-1 because the concentration of A is increased 4 Times. In third blank space concentration of A will be 0.1 mol L^-1 because the rate is same as in experiment I.