The rate of the following reaction:
Radioactive decay of 90Sr,
is characterized by the rate constant k = 0.024 year-1. What is the half-life (in years) for this reaction at the same conditions when the initial concentrations are 0.0356 M? (Round your answer to 3 significant figures.)
Hint: The units of the rate constant are linked with the order of reaction.
Answer:________________years
Radioactive decay is a first order reaction. t1/2 = 0.693/k
Solve for t1/2.
It doesn't matter that conditions are changed, the half-life is the same; half life of first order reactions don't depend upon the initial concn. You can show that, however, by using
ln(No/N) = kt
Substitute 0.0356 M for No and half that for N and 0.024 for k. t1/2 will be the same as obtained by 0.693/0.024.
To determine the half-life of a reaction, you need to use the rate constant and the initial concentration of the reactant. In this case, the rate constant (k) is given as 0.024 year⁻¹, and the initial concentration (A₀) is 0.0356 M.
The half-life (t₁/₂) is the time it takes for the concentration of the reactant to decrease to half its initial value. It can be calculated using the formula:
t₁/₂ = ln(2) / k
Where ln is the natural logarithm, and k is the rate constant.
Plugging in the given values:
t₁/₂ = ln(2) / 0.024
Using a calculator, evaluate the natural logarithm of 2 (ln(2)), and then divide it by 0.024 to find the value of t₁/₂. Make sure to round your answer to 3 significant figures as requested.
Calculating the value of t₁/₂ will give you the half-life of the reaction in years.