Given information:

  dxdt=K(a−x)(b−x)12where x=[HBr] , a and b are the initial concentration of hydrogen and bromine.

  a=b

Formula used :

  ∫xndx=xn+1n+1+C

Calculation:

When a=b

  dxdt=K(a−x)(a−x)12⇒dxdt=K(a−x)32⇒dx(a−x)32=Kdt

Integrating both sides, we get,

  ∫(a−x)−32dx=K∫dt+C         [C is integrating constant]⇒112(a−x)−12=Kt+C⇒2(a−x)=Kt+C

Given that x(0)=0 , then

  2(a−0)=K×0+C⇒C=2a

So, substituting the value of C , we get,

  2(a−x)=Kt+2a⇒(a−x)12=2Kt+2a⇒(a−x)=(2Kt+2a)2⇒x=a−4a(aKt+2)2

  x(t)=a−4a(aKt+2)2

Given information:

  dxdt=K(a−x)(b−x)12   ..........(i)

  a>b

Formula Used:

  ∫dxa2+x2=1atan−1xa+C

Calculation:

  dxdt=K(a−x)(b−x)12   ..........(i)

We substitute u=b−a

  ⇒2ududt=−dxdt

Therefore, equation (i) can be written as,

  −2ududt=Ku(a−b+u2)           [u2=b−xu2−b=−x⇒a−b+u2=a−x]

  ⇒dudy=−K2(a−b+u2)⇒du(a−b)+u2=−K2dt

Integrating both side, we get,

  1a−btan−1ua−b=−K2t+Ca−b                 [Ca−b is integrating constant]⇒tan−1ua−b=(−K2a−b)t+C⇒tan−1b−xa−b=−K2(a−b)t+C⇒K2(a−b)t=C−tan−1b−xa−b⇒t=2CKa−b−2Ka−btan−1b−xa−b⇒t=D−2Ka−btan−1b−xa−b                [D=2CKa−b]