for reaction 2A + B2------2AB . following data was obtained
S.No [A] [B2] rate
1 0.015 0.15 1.8 x 10-3
2 0.09 0.15 1.08 x 10-2
3 0.015 0.45 5.4 x 10-3
calculate the rate constant and rate of formation of AB when [A] is 0.02 and [B2] is 0.04 mol lt-1 at 300K
Rate of reaction = K[A]m[B]n
Where m and n are order with respect to A and B
Rate of disappearance of B2 = k[A]m[B]n
r1 =1.8*10-3 = k[0.015]m[0.15]n
r2=1.08*10-2 =k[0.09]m[0.15]n
r3=5.4*10-3 =k[0.15]m[0.45]n
dividing r1/r2
(1.8*10-3/1.08*10-2)=k[0.015/0.09]m
thus we get m=1
similarly dividing r1/r3 we get n=1
Rate of disappearance of B2 =k[0.015]1[0.15]1
1.8*10-3=k[0.015]1[0.15]1
Rate constant (k) = 0.8 litre mol-1 time-1
rate of formation of AB
1/2 d[AB]/dt =-d[B2]/dt
d[AB]/dt = 2*(-d[B2]/dt)
=2*k[A]1[B]1
=2*0.8*(0.02)1*(0.04)1
= 1.28*10-3
Thus rate of formation of [AB] =1.28*10-3
Where m and n are order with respect to A and B
Rate of disappearance of B2 = k[A]m[B]n
r1 =1.8*10-3 = k[0.015]m[0.15]n
r2=1.08*10-2 =k[0.09]m[0.15]n
r3=5.4*10-3 =k[0.15]m[0.45]n
dividing r1/r2
(1.8*10-3/1.08*10-2)=k[0.015/0.09]m
thus we get m=1
similarly dividing r1/r3 we get n=1
Rate of disappearance of B2 =k[0.015]1[0.15]1
1.8*10-3=k[0.015]1[0.15]1
Rate constant (k) = 0.8 litre mol-1 time-1
rate of formation of AB
1/2 d[AB]/dt =-d[B2]/dt
d[AB]/dt = 2*(-d[B2]/dt)
=2*k[A]1[B]1
=2*0.8*(0.02)1*(0.04)1
= 1.28*10-3
Thus rate of formation of [AB] =1.28*10-3