The first order integrated rate equation is

Options

A. `k=(x)/(t)`
B. `k=-(2.303)/(t)"log"(a)/(a-x)`
C. `k=(1)/(t)In(a)/(a-x)`
D. `k=(1)/(t)(x)/(a(a-x))`

Answer

C

Explanation

For first order, consider a reaction
` R rarr " Products " `
`" rate " = (d[R])/(dt) = k [R] `
or `(d[R])/([R]) = kdt `
On intergrating Eq. (i)
`int (d[R])/([R])= k intdt `
In [R]=-kt +C
`At t=0 [R] =[R_0]` [ Where , R concentration at time , t i.e a-x and `R_0` is the initial concentration i.e a ]
`" In" [R_0]=C `
On putting the value of C in Eq, (ii) we, get
`"In" [R] =-kt + "In"[R_0\]`
`-kt =In [R]-"In "[R_0] " and " kt ="In" [R_0]`
or `K=1/tIn ([R_0])/([R])=1/t "In"a/(a-x)`