The isotope sodium-22 has a half life of 2.60 years. What is the 1st order rate constant in inverse years?

I am using the Ln (A/A0) = -Kt formula. Is that correct?

Yes, that is right if yu use it correctly. To use it, just substitute 100 for Ao and 50 for A and solve for t which then will be t1/2 Actually it doesn't matter what number you pick for Ao as long as A is 1/2 that. Another way and slightly shorter is to tkae ln 1/2 (which is 0.693), then

k = 0.693/t1/2

Thank you!!!

Yes, the formula you mentioned, Ln(A/A0) = -Kt, is correct for a first-order reaction. In this formula, A is the amount of the isotope remaining at a given time (t), A0 is the initial amount of the isotope, K is the first-order rate constant, and Ln is the natural logarithm.

To determine the first-order rate constant (K) in inverse years for the given isotope (sodium-22), you can rearrange the equation as follows:

Ln(A/A0) = -Kt

Rearranging the equation, we get:

K = -Ln(A/A0) / t

Given that the half-life of sodium-22 is 2.60 years, we can use this information to find the value of A/A0 at t = 2.60 years. In a first-order reaction, the amount remaining (A) at the half-life is equal to half of the initial amount (A0), so A/A0 would be 1/2, or 0.5.

Therefore, we can substitute the values into the equation:

K = -Ln(0.5) / 2.60

Using a scientific calculator to evaluate the natural logarithm of 0.5, we get approximately Ln(0.5) ≈ -0.693.

Substituting the values, we have:

K ≈ -(-0.693) / 2.60

Simplifying, we get:

K ≈ 0.2665 inverse years

Hence, the first-order rate constant for sodium-22 is approximately 0.2665 inverse years.