Question

the specific rate (in mol L^−1s^−1) of a first order reaction having half life of 6.93 seconds

Answer 1

Answer : The specific rate (in mol L^−1s^−1) of a first order reaction is 0.1

Explanation :

Specific rate of a reaction is the rate when the concentration of reactants is equal to 1 M.

The rate equation for first order reaction can be written as

$Rate = k [A]$

Where, k = Rate constant

[A] = concentration of reactant A

We know that [A] = 1 mol/L

Let us find k using the given half life.

The relationship between k and t(1/2) is given by the following equation.

$k = \frac{0.693}{t_{1/2}}$

We have been given t(1/2) = 6.93 s

Let us substitute this in the above equation.

$k = \frac{0.693}{6.93s}$

$k = 0.1 s^{-1}$

The specific rate constant is 0.1 s⁻¹.

Let us substitute this value of k in rate equation.

$Rate = 0.1s^{-1}\times 1mol. L^{-1}$

Thereofore , Rate = 0.1 mol L⁻¹ s⁻¹.