Question

For a gaseous phase reaction
2A + B2 → 2AB the following rate data was obtained at 300K ​

Answer 1

Answer:

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)

Hence rate of formation of [AB] =1.28*10^(-3 )

Answer 2

Answer:

0.8 L/mol sec

Explanation:

rate =K[A]^m[B]^n

rate 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[0.015]m[0.45]n

divide R1by R2

=1.8×10^-3/1.08×10^-2=[0.015/0.09]m

m= 1

similarly R2 divide by R3

we get n=1

now put these values in equation

B2=K[0.015]^1[0.15]^1

K=1.8×10^-3/0.015×0.15

K= 1.8×10^-3/2.25×10^-3

K=0.8 L/mol sec