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.