Answer :
The first-order rate constant for this deactivation is 0.052 min⁻¹ .Using the equation for first-order kinetics:
[tex]ln(A_{}/A) = kt[/tex] .Where A₀ is the initial concentration, A is the concentration at time t, k is the first-order rate constant, and t is time.
We can convert these percentages into decimal fractions by dividing them by 100.
Using the given data, we can calculate the values [tex]ln(A_{0}/A)[/tex]of for different time intervals:
[tex]ln(1.00/0.90) = 0.105[/tex]
[tex]ln(1.50/0.968) = 0.422[/tex]
[tex]ln(2.00/0.990) = 0.203[/tex]
[tex]ln(2.50/0.997) = 0.00993[/tex]
[tex]ln(3.00/0.999) = 0.00100[/tex]
Now, we can plot ln(A₀/A) vs. time (t) and get a straight line with a slope equal to -k, the first-order rate constant. The slope of the line can be calculated using any two points on the line:
[tex]slope =(ln(A_{0}/A_{2}) - ln(A_{0}/A_{1}))/(t_{2} - t_{1})[/tex]
Using the values for ln(A₀/A) and time from the given data, we get:
[tex]slope =(0.00100 - 0.105)/(3.00 - 1.00) = -0.052[/tex]
Therefore, the first-order rate constant (k) for this deactivation is:
k = 0.052 min⁻¹ (rounded to three significant figures)
Learn more about first-order kinetics here:
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