A first order gaseous reactions has \(K=1.5 \times 10^{-0} \mathrm{sec}^{-1}\) at \(200^{\circ} \mathrm{C}\). If the reaction is allowed to run for 10 hour, what percentage of initial concentration would have changed into products. What is the half life period of reaction?

In the realm of chemistry, the rate constant plays a pivotal role in understanding how quickly a reaction proceeds. It is denoted by the symbol 'k' and is a measure of the speed at which reactants are converted into products.

For a first-order reaction, the rate constant is especially crucial because it directly correlates with the lifespan of the reactants. It tells us how rapidly the concentration of the reactant will decrease over time. The units of the rate constant for a first-order reaction are reciprocal seconds (\text{s}^{-1}), indicating how many times per second the reaction process is happening.

In our exercise, the given rate constant is \(1.5 \times 10^0 \text{s}^{-1}\), which is a relatively high value, suggesting that the reaction proceeds quickly. Understanding this value is fundamental to calculating either the concentration of reactants at a certain time or the half-life of the reaction.

Calculating the concentration of reactants or products during a chemical reaction involves using the first-order reaction equation \([A]_t = [A]_0 e^{-kt}\).

This equation is used to determine the concentration of a reactant \([A]_t\) at any time 't' during the reaction. \([A]_0\) is the initial concentration and 'k' is our rate constant. The exponential component comes into play because as the reaction proceeds, the rate at which the concentration of the reactant decreases is proportional to its current concentration.

To apply the equation from our exercise, after converting the time from hours to seconds, we plugged in the values into the equation to find the remaining concentration after 10 hours. This gives us insight into how much of the starting material remains at a given point in time, critical for both theoretical understanding and practical applications, such as in pharmaceuticals where the concentration of a drug in the bloodstream may be modeled using these equations.

The half-life of a chemical reaction is the time required for the concentration of a reactant to decrease to half of its initial value. In the context of a first-order reaction, the half-life is independent of the initial concentration and can be found using the formula \(t_{1/2} = \frac{\ln(2)}{k}\), where \(k\) is the rate constant.

This expression arises because, for a first-order reaction, the time it takes for the concentration to reduce by half remains the same throughout the process. For the scenario provided in the exercise, the half-life calculation is straightforward and gives a singular value.

Knowing the half-life can be very useful, for instance, in the pharmaceutical industry for determining how frequently a medication should be administered to maintain its therapeutic level or in environmental science to estimate how long it will take for pollutants to decrease in concentration.

Exponential decay describes the process by which a quantity decreases at a rate proportional to its current value, which in chemistry refers to the concentration of a reactant in a first-order reaction. This type of decay is described by the exponential function seen in the equation \([A]_t = [A]_0 e^{-kt}\).

Exponential decay is characterized by the presence of 'e', the base of natural logarithms, and is a fundamental concept illustrating how substances transform over time. It’s everywhere in nature, from radioactive decay to the decrease of a drug’s concentration in the bloodstream.

In the given exercise, applying the concept of exponential decay allows us to calculate how much of the reactant has been transformed into products after a certain period. The use of the natural exponential function also makes it simpler to calculate related parameters like half-life, as natural logarithms are inherently related to the base 'e'.