for reaction 2A + B_{2}------2AB . following data was obtained

S.No [A] [B_{2}] 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 [B_{2}] is 0.04 mol lt^{-1} at 300K

^{m}[B]

^{n}

Where m and n are order with respect to A and B

Rate of disappearance of B

_{2}= k[A]

^{m}[B]

^{n}

r

_{1}=1.8*10

^{-3}= k[0.015]

^{m}[0.15]

^{n}

r

_{2}=1.08*10

^{-2}=k[0.09]

^{m}[0.15]

^{n}

r

_{3}=5.4*10

^{-3}=k[0.15]

^{m}[0.45]

^{n}

dividing r

_{1}/r

_{2}

(1.8*10

^{-3}/1.08*10

^{-2})=k[0.015/0.09]

^{m}

thus we get m=1

similarly dividing r

_{1}/r

_{3}we get n=1

Rate of disappearance of B

_{2}=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[B

_{2}]/dt

d[AB]/dt = 2*(-d[B

_{2}]/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}

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