Question

In a chemical reaction of a given substance being converted into another at a rate propositional to the amount of substance unconverted .If 1/5 th of the original amount transform in 4minuets. How much time will be required to transform half of original amount.​

Answer 1

t = $\frac{ln\frac{1}{2} }{-(\frac{1}{4})*ln(\frac{4}{5})}$

Explanation:

The given problem describes a first-order chemical reaction, where the rate of transformation is directly proportional to the amount of substance unconverted.

Let's denote the original amount of substance as "A" and the time taken for half of the original amount to transform as "t". According to the problem, 1/5th of the original amount transforms in 4 minutes.

We can set up the following equation based on the first-order reaction kinetics:

A(t) = A(0) * $e^{-kt}$

where A(t) is the amount of substance remaining at time "t", A(0) is the initial amount, and "k" is the rate constant. Since 1/5th of the original amount transforms, $\frac{A(t)}{A(0)}$ = $\frac{4}{5}$.

Substituting the given values, we have:

($\frac{4}{5}$) = $e^{-4k}$.

To find the rate constant "k", we take the natural logarithm (ln) on both sides:

ln($\frac{4}{5}$) = -4k.

Solving for "k", we find k = -($\frac{1}{4}$) * ln($\frac{4}{5}$).

Now, we can determine the time required to transform half of the original amount. Since we want $\frac{A(t)}{A(0)}$ = $\frac{1}{2}$, we substitute these values into the equation:

($\frac{1}{2}$) = $e^{-kt}$

Plugging in the value of "k" we previously found, we solve for "t":

($\frac{1}{2}$) = $e^{-((-\frac{1}{4})*ln(\frac{4}{5})*t) }$

Taking the natural logarithm on both sides:

ln($\frac{1}{2}$) = $-((-\frac{1}{4})*ln(\frac{4}{5})*t) }$

Simplifying further:

t = $\frac{ln\frac{1}{2} }{-(\frac{1}{4})*ln(\frac{4}{5})}$

Using the properties of logarithms and evaluating the expression, we can find the value of "t".

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