Question

A second-order reaction has a half-life of 12 s when the initial concentration of reactant is 0.98 M. The rate constant for this reaction is __________ M-1s-1.

Answer 1

Answer:

0.085 (Ms)⁻¹

Explanation:

Half life = 12 s

$[A_o]$ is the initial concentration = 0.98 M

Half life expression for second order kinetic is:

$t_{1/2}=\frac{1}{k[A_o]}$

So,

$12=\frac{1}{k\times 0.98}$

$x=\frac{25}{294}$

k = 0.085 (Ms)⁻¹

The rate constant for this reaction is 0.085 (Ms)⁻¹ .

Answer 2

The rate constant for the reaction will be "0.085 M⁻¹s⁻¹".

The given values:

• Half life = 12 s
• Initial concentration, $[A_o]$ = 0.98 M

As we know the formula,

→ $t_{\frac{1}{2} } = \frac{1}{k[A_o]}$

By substituting the given values, we get

  $12 = \frac{1}{k\times 0.98}$

   $k = \frac{25}{294}$

      $= 0.085 \ M^{-1}s^{-1}$

Thus the above answer is appropriate.

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