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What is the half-life (in seconds) of a zero-order reaction which has an initial reactant concentration of 0.884 M with a k value of 5.42 × 10–2 M/s?

Respuesta :

Edufirst
Answer: 8.15s


Explanation:


1) A first order reaction is that whose rate is proportional to the concenration of the reactant:


r = k [N]


r = - d[N]/dt =


=> -d[N]/dt = k [N]


2) When you integrate you get:


N - No = - kt


3) Half life => N = No / 2, t = t'


=> No - No/ 2 = kt' => No /2 = kt' => t' = (No/2) / k


3) Plug in the data given: No = 0.884M, and k = 5.42x10⁻²M/s


t' = (0.884M/2) / (5.42x10⁻²M/s) = 8.15s

Answer:

The half-life (in seconds) of a zero-order reaction is 8.15 seconds.

Explanation:

Initial concentration of the of the reactant = [tex][A_o]=0.884 M[/tex]

The value of rate constant = [tex]k=5.42\times 10^{-2} M/s[/tex]

The half life for zero order reaction is given as:

[tex]t-{\frac{1}{2}}=\frac{[A_o]}{2k}=\frac{0.884 M}{2\times 5.42\times 10^{-2} M/s}=8.15 s[/tex]

The half-life (in seconds) of a zero-order reaction is 8.15 seconds.