Animated Solution for Chemistry - Chemical Kinetics: For a certain first order reaction 32% of the reactant is left after 570 s. The rate constant of this reaction is ...... $\times 10^{-3}{\text{ s}}^{-1}$. (Round off to the nearest integer). [Given, ${\log }_{10}2=0.301$, $\ln 10=2.303$]

Imagine a radioactive sample or a chemical reactant decaying over time. For a first-order reaction, the concentration drops exponentially, never quite reaching zero. We start our observation at time $t=0$. At this initial moment, we have $100\%$ of our reactant present. This is our starting point, our initial concentration $[A{]}_0$.

As time passes, the concentration steadily decreases. The problem states that after $570\text{ s}$, only $32\%$ of the reactant is left. The mathematical tool that connects these two states is the integrated rate law for a first-order reaction. This equation is the heart of chemical kinetics.

The integrated rate law for a first-order reaction is given by:

Let's substitute our known values into this powerful equation. Our initial concentration, $[A{]}_0$, is $100$. The final concentration, $[A{]}_t$, is $32$. And the time elapsed is $570\text{ s}$.

Here is where we need to be careful. We have a natural logarithm ($\ln$) in our equation, but the problem gives us the value of ${\log }_{10}$. We cannot mix these up! So, we must convert the natural log to base ten. We do this by multiplying the expression by the conversion factor, $2.303$.

Don't rush to divide $100$ by $32$. That will give you a messy decimal. Instead, use the elegant properties of logarithms! We know that $\log \left(\frac{a}{b}\right)=\log (a)-\log (b)$.

Furthermore, ${\log }_{10}(100)$ is exactly $2$. And $32$ is $2^5$, so ${\log }_{10}(32)$ becomes $5{\log }_{10}(2)$. See how beautifully it simplifies?

We are given that ${\log }_{10}(2)=0.301$. Let's plug that specific value in.

By using log properties, we completely avoided any complex division!

Now, let's bring this result back to our main rate constant equation. We have $2.303$ multiplied by $0.495$, all divided by $570$.

The numerator multiplies out to approximately $1.14$. And $1.14$ divided by $570$ is exactly $0.002$, or $2\times 10^{-3}{\text{ s}}^{-1}$!

Comparing our final result with the format given in the question ($......\times 10^{-3}{\text{ s}}^{-1}$), the integer value we need to fill in the blank is $2$. We have successfully solved the problem!