Question

expression for the rate constant of the first order reaction

Answer 1

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Answer 2

Answer: The expression for first order kinetics is given below.

Explanation:

Equation for the first order kinetics is given as:

$A\rightarrow \text{Products}$

Let us start with 'a' moles per liter of the reactant A. After time 't', suppose 'x' moles per liter of it have decomposed. Thus, the concentration of A after time 't' will be (a - x) moles per liter.

Now, according to Law of mass Action,

Rate of the reaction is directly proportional to the concentration of reactant after time 't'

$\text{Rate of reaction}\propto (a-x)$

Or,

$\frac{dx}{dt}\propto (a-x)$

Removing the proportionality sign, we get:

$\frac{dx}{dt}=k(a-x)$       .....(1)

where k = rate constant for first order

The expression for 'k' can be derived as from equation 1:

$\frac{dx}{(a-x)}=k.dt$       .....(2)

Integrating equation 2, we get:

$\int \frac{dx}{(a-x)}=\int k.dt$

$-ln(a-x)=kt+I$       .....(3)

I = constant of integration

At the start, when t = 0, x = 0 (as no substance has decomposed at the start of the reaction)

Putting the values in equation 3, we get:

$-ln(a-0)=k\times 0+I$

$-ln(a)=I$       ....(4)

Substituting the value of 'I' in equation 3, we get:

$-ln(a-x)=kt+(-ln(a))\\\\kt=ln(\frac{a}{a-x})$    ......(5)

$k=\frac{1}{t}ln(\frac{a}{a-x})\\\\k=\frac{2.303}{t}\log \frac{a}{a-x}$   .....(6)

If the initial concentration is $[A]_o$ and the concentration after time 't' is [A], then putting $a=[A]_o$ and $(a-x)=[A]$, equation 6 becomes:

$k=\frac{2.303}{t}\log \frac{[A]_o}{A}$

Hence, the expression for first order kinetics is given above.