Question

A find order reaction is found to have a rate constant k = 7.39×10^-5 Sec ^-1. calculate its two third life​

Answer 1

Answer:

=0.693k=0.6937.39×10−5s−1=9.3775×103s

=0.0938*10^5s

=9.38*10^3s

Answer 2

Two-third life is 1.49 × $10^{9}$ sec.

• A chemical reaction that has a reaction rate that is linearly dependent on the concentration of just one ingredient is known as a first-order reaction. Precisely, a first-order reaction is a chemical reaction in which only one of the reactants' concentrations changes and the rate of the reaction changes as a result.
• A chemical process can be explained at the molecular level using a differential rate rule.
• The value of a reaction's rate constant can be determined empirically using integrated rate equations.

Here, according to the given information, we are given that,

A first order reaction is given and we need to find its two-third life.

Now, we have $A_{0}$ as a, say.

Then, after $\frac{2}{3}$rd of its life, we get, A is equal to,

$a-\frac{2}{3} a = \frac{3a-2a}{3} =\frac{a}{3} .$

Now, we know that,

$t_{n} =\frac{2.303}{k}log\frac{A_{0} }{A}$

Then, we get for n = $\frac{2}{3} ,$

$t_{\frac{2}{3} } =\frac{(2.303).(100000)}{7.39}log\frac{3a }{a}=14868.88 =$ 1.49 × $10^{9}$ sec.

Hence, two-third life is 1.49 × $10^{9}$ sec.

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