The half-life of a first order reaction is determined to be 92.0 years. How long will it take for the concentration of the reactant to reach 2% of its initial value?
You meant to say the question is: The half-life of a first order reaction is determined to be 85.5 years. How long will it take for the concentration of the reactant to reach 2% of its initial value?
Make use of the formula t=2.303/k*log[Ri/Rf]....where k=0.639/halflyf Rf=2/100Ri..
Got it 519.23512
92.0 years X (multiply) with 5.64386
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To solve this problem, we need to use the formula for the half-life of a first-order reaction:
t1/2 = (0.693 / k)
where t1/2 is the half-life of the reaction, and k is the rate constant.
First, we need to find the rate constant (k). We can use the half-life of the reaction in years to do this:
t1/2 = 92.0 years
Rearranging the formula, we have:
k = 0.693 / t1/2
Plugging in the value for t1/2, we get:
k = 0.693 / 92.0 years
Next, we can use the rate constant to find the time it takes for the concentration to reach 2% of its initial value.
The integrated rate law for a first-order reaction is given by:
ln(C / Co) = -kt
where C is the concentration at time t and Co is the initial concentration.
In this case, we want to find the time it takes for the concentration (C) to reach 2% of its initial value (Co), so we can set up the equation:
ln(0.02) = -kt
Substituting the value of k we found earlier, we have:
ln(0.02) = -[(0.693 / 92.0 years) * t]
Now, solve for t by rearranging the equation:
t = ln(0.02) / (-(0.693 / 92.0 years))
Evaluate the expression on the right side of the equation:
t = ln(0.02) / (-0.00752)
Using a calculator, ln(0.02) ≈ -3.912 and -0.00752 ≈ -0.008.
Therefore, we have:
t = -3.912 / -0.008
t ≈ 489 years
So, it will take approximately 489 years for the concentration of the reactant to reach 2% of its initial value.