Find the ratio for the following for first order reaction t87. 5%/t50%=

To find the ratio of t_87.5% to t_50% for a first-order reaction, we can use the properties of first-order kinetics. In first-order reactions, the time it takes for a certain percentage of reactant to be consumed can be calculated using the half-life and the natural logarithm. Let's break this down step by step.

Understanding First-Order Reactions

In a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant. This means that as the concentration decreases, the rate of reaction also decreases. The half-life (t_1/2) of a first-order reaction is constant and is given by the formula:

t_1/2 = 0.693/k

where k is the rate constant of the reaction.

Calculating t_50% and t_87.5%

For a first-order reaction, the time required to reach a certain percentage of completion can be determined using the following relationship:

t = (1/k) * ln([A₀]/[A])

Where [A₀] is the initial concentration and [A] is the concentration at time t.

• For t_50%, we have [A] = 0.5[A₀]. Thus:

t_50% = (1/k) * ln(2) = 0.693/k

• For t_87.5%, we have [A] = 0.125[A₀]. Thus:

t_87.5% = (1/k) * ln(8) = (1/k) * 2.079 = 2.079/k

Finding the Ratio

Now that we have both t_50% and t_87.5%, we can find the ratio:

Ratio = t_87.5% / t_50%

= (2.079/k) / (0.693/k)

= 2.079 / 0.693

≈ 3.00

Final Thoughts

Thus, the ratio of t_87.5% to t_50% for a first-order reaction is approximately 3. This means that it takes about three times longer to reach 87.5% completion of the reaction compared to reaching 50% completion. This relationship highlights the exponential nature of first-order kinetics, where the time required increases significantly as the reaction approaches completion.