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2 years ago

10

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 valu

e of 5.42 × 10–2 M/s?

2 answers:

pantera1 [17]2 years ago

6 0

<span>Answer: 8.15s
</span><span />

<span>Explanation:
</span><span />

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

<span>r = k [N]
</span><span />

<span>r = - d[N]/dt =
</span><span />

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

<span>2) When you integrate you get:
</span><span />

<span>N - No = - kt
</span>
<span></span><span /><span>
3) Half life => N = No / 2, t = t'
</span><span />

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

<span>3) Plug in the data given: No = 0.884M, and k = 5.42x10⁻²M/s
</span>
<span /><span /><span>
t' = (0.884M/2) / (5.42x10⁻²M/s) = 8.15s</span>

lorasvet [3.4K]2 years ago

5 0

Answer:

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

Explanation:

Initial concentration of the of the reactant = $[A_o]=0.884 M$

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

The half life for zero order reaction is given as:

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

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

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