Chemistry, asked by khushirathore37, 8 months ago

derive the rate law equation k=2.303/t log a/a-x for first order equation and establish the relation between half life and rate constant​

Answers

Answered by ritiksingh83032830
3

Explanation:

Consider a first order reaction A —> products; velocity of the reaction is given by,

dx/dt = k [A]

Let the initial concentration of the reactant be ‘a’ mole / dm3. Let, ‘x’ mole/dm3 decompose in‘t’ seconds. The remaining concentration of the reactant is ‘a-x’ mole/dm3.

Therefore, the velocity of the reaction after time‘t’ is given by

dx/dt = k [a-x]

Rearranging,

dx/dt = k [a-x]

Integrating the above expression,

∫ dx/[a-x] = ∫ k.dt

-ln (a-x) = kt + C --- (1)

where C is an integration constant.

When t = 0, x = 0 or a-x = a

Equation (1) becomes,

-ln a = 0 + C or C = -ln a

Substituting this in equation (1),

-ln (a-x) = kt – ln a

kt = ln a – ln (a-x)

= ln (a/a-x)

= 2.303 log (a/a-x) [ ln = 2.303 log]

k = 2.303/t log (a/a-x)

This is the rate equation for a first order reaction.

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