Ncert Solutions Chemistry Class 12th

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.

4.20 By comparing Experiment I and IV if we increase the concentration of A by 4 times then Rate also increased by 4 times. That means order with respect to A is 1.

By comparing Experiment II and III if we double the concentration of B Rate increases by 4 times that means order with respect to B is 2.

Rate law of reaction will be, Rate = k [A] [B]²

To find K, K = rate/ [A] [B]2 i.e. K = 6.0 * 10^-3/ [0.1] [0.1]²

K = 6 mol^-2L²sec^-1

Order with respect to A and B is 1 and 2 respectively. And value of K (rate constant) is = 6 mol^-2L²sec-1

4.19 When concentration of B is changed then rate of reaction doesn't change that means order with respect to B is 0. But when the concentration of A is doubled rate increased by 2.82 times i.e.2^1.5 = 2.82.Hence order with respect to A is 1.5.

Order with respect to A and B is 1.5 and 0, respectively.

4.18  (i) Order is power raised to reactant in rate law, hence,

Rate = k [A] [B]²

(ii) When the concentration of B is increased three times then the rate is affected by the square of The rate is increased by 9 Times.

(iii) When the concentration of reactant both A and B is doubled then the rate will have affected as square of reactant B and Two times of Reactant Overall increase in rate is 8 times

(a) When the concentration of B is increased by three times, the rate is increased by nine times

(b) When of both reactants is doubled, the Rate increases 8 times.