Question

A certain first-order reaction a → b is 25% complete in 42 min at 25°c. What is the half-life of the reaction?

Answer 1

Answer:

$t_{\frac{1}{2}} = 101.16$ minutes

Explanation:

For a First order reaction, the rate constant is determined by

$k = \frac{2.303}{t} log \frac{a}{a-x}$

Where

k= rate constant

t= time required for a reaction

a= initial concentration

x= amount of reactant decomposed in time t

a-x= amount of reactant left after time t

Let the initial concentration be $100$

Then

$x = 25\\a-x = 75$

Substituting the given values in above equation, we get -

$k = \frac{2.303}{42} log \frac{100}{75}$

$k = 6.85 *10^{-3}$

Also,

$k = \frac{0.693}{t_{\frac{1}{2}} } \\t_{\frac{1}{2}} = \frac{0.693}{6.85 *10^{-3}}$

$t_{\frac{1}{2}} = 101.16$ minutes

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Answer 2

First-order reaction is defined as the reaction in which the rate of the reaction is dependent on the concentration of only one reactant.

The half-life of the reaction is 101.6 minutes.

The rate constant for the first-order reaction can be determined by:

$\text k &=\dfrac{2.303}{\text t}\;\text{log}\;\dfrac{\text a}{\text a-x}$

In which:

x= amount of reactant decomposed in time t

k= rate constant  

a-x= amount of reactant left after time t

t= time required for a reaction

a= initial concentration

The initial concentration of the reaction is 100.

Now, given:

x = 25

a - x = 75

t = 42

Putting the values in the above equation, we get:

$\text k &=\dfrac{2.303}{\text t}\;\text{log}\;\dfrac{\text a}{\text a-x}\\\\\text k &=\dfrac{2.303}{42}\;\text{log}\;\dfrac{100}{75}$

k = 6.85 x 10 ⁻³

The half-life will be:

$\begin{aligned} \text k = \dfrac{0.693}{\text t_\frac{1}{2}}\\\text t_\frac{1}{2} = \dfrac{0.693}6.85 \times 10^3}\\\\\\\text t_\frac{1}{2} = 101.6 \end{aligned}$minutes.

Therefore, the half-life of the reaction is 101.6 minutes.

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