Answers

Assignment No – 1

Chemical Kinetics
Ans 1. Zero order reaction.

Ans 2: Rate = – change in conc. of A/2 x time interval
= – [0.4‐0.5]/2X10 = 0.005 mol L–1min–1

Ans 3: Given = 60 s–1, [R] =1, [R]=1/16, t =?
t = 2.303/k log[R]o /[R]
= 2.303/60 log 16 = 3.84X10–2 s

Ans 4. When reaction is completed 99.9%, [R]n = [R]0 – 0.999[R]0
k = 2.303/t log R0/Rn
= 2.303/t log[R]0 /[R]0 – 0.999 [R]0
= 2.303/t log103
t = 6.909/k
For half‐life of the reaction t1/2 = 0.693/k
t/t1/2 =6.909/0.693 =10

Ans 5. Log k2/k1=Ea/2.303R log[T2 – T1/T2T1]
Log 0.07/0.02 =Ea/2.303X8.314X log[700‐500/700X500]
Ea = 18230.8 J
log k = log A – Ea/2.303RT
log 0.02 = log A – 18230.8/20303X8.314X500
A= 1.61

Ans 6. Given t =40 min, [R]o = 100, [R]=100 - 30=70
k = 2.303/t log R0/R
= 2.303/40 log100/70
=0.0575(log100‐log70) = 0.0575(2‐1.84) = 0.00890 min–1
t1/2 = 0.693/k = 0.693/0.00890 = 77.86 min

Ans 7. (a) Given t1/2= 5730 years, [R]o = 100, [R] = 80
K = 0.693/t = 0.693/5730 = 1.21X10–4 year–1
t =2.303/K log[R]o/[R] = 2.303/1.21X10 log100/80 = 1.9033X10-4(log100–log80)
= 1.9033X 10–4 (2‐1.9031) = 1.9033X 10–4 (0.0969) = 1845 year
(b) (i) Rate = K[A]1 [B]2
(ii) Rate = K[A]1 [3B]2 = 9K[A]1 [B]2 hence, it becomes 9 times.
(iii)Rate = K[2A]1 2[B]2 = 8K[A]1 [B]2 hence, it becomes 8 times.


Que 8.  MCQS
1. (iii) 

2.(iii) 

3. (iii) 

4. (i) 

5. (ii)       

6. (iii)                 

7. (iii) 

8. (i), (iv)

9. (i), (ii) 

10. (i), (iv)