Answer:
The reaction is second-order, and k = 0.0267 L mol^-1 s^-1
Explanation:
<u>Step 1:</u> Data given
The initial concentration is 0.331 M
half‑life time = 113 s
The same reactant decomposes with a half‑life of 243 s when its initial concentration is 0.154 M.
<u>Step 2: </u>Determine the order
The reaction is not first-order because the half-life of a first-order reaction is independent of the initial concentration:
t½ = (ln(2))/k
Calculate k for the two conditions given:
⇒ 113 s with initial concentration is 0.331 M
t½ = ([A]0)/2k
113 s = (0.331 M)/2k
k = 0.00146 mol L^-1 s^-1
⇒ 243 s with an initial concentration is 0.154 M
t½ = ([A]0)/2k
243 s = (0.154 M)/2k
k = 0.000317 mol L^-1 s^-1
The <u>values of k are different</u>, so that rules out zero-order.
<u>Step 3: </u>Calculate if it's a second-order reaction
For a second-order reaction, the half-life is given by the expression
t½ = 1/((k*)[A]0))
<u>Calculate k for the two conditions given:
</u>
⇒ 113 s when its initial concentration is 0.331 M
t½ = 1/((k*)[A]0))
113 s = 1/(k*(0.331 M))
k = 1/((0.331 M)*(113 s)) = 0.0267 L mol^-1 s^-1
⇒ 243 s when its initial concentration is 0.154 M
t½ = 1/((k*)[A]0))
243 s = 1/(k*(0.154 M))
k = 1/((0.154 M)*(243 s)) = 0.0267 L mol^-1 s^-1
The values of k are the same, so the reaction is second-order, and k = 0.0267 L mol^-1 s^-1
