Respuesta :
Answer:
a) Half life = 2.56 s
b) Remaining concentration after 5.16 s = 0.0128 M
Explanation:
Given,
Rate constant = [tex]0.271\;s^-1[/tex]
a)
Half life of a first order reaction is given by,
Half life = [tex]\frac{ln2}{rate\;constant}[/tex]
= [tex]\frac{ln2}{0.271} =2.56s[/tex]
b)
Integrated rate law for first order reaction is:
[tex]A = A_0\times exp(-kt)[/tex]
Where,
A = Concetration after time 't'
A0 = Initial concentration = 0.052 M
k = rate constant = 0.271 s^-1
t = time = 5.16 s
[tex]A = A_0\times exp(-kt)[/tex]
[tex]A = 0.52\times exp(-0.271 \times 5.16)\\\; \; \; \; \; =0.0128\;M[/tex]
a) Half life = 2.56 s
b) Remaining concentration after 5.16 s = 0.0128 M
Half life for first order reaction:
The half-life of a first-order reaction is a constant that is related to the rate constant for the reaction:
[tex]t_{1/2}=\frac{0.693}{k}[/tex]
Given:
Rate constant = [tex]0.271 s^{-1}[/tex]
a)
[tex]\text{Half life} =\frac{0.693}{k}\\\\ t_{1/2}= \frac{0.693}{0.271}\\\\ t_{1/2}= 2.56s[/tex]
b)
Rate law for first order reaction is given by:
[tex]A=A_0*exp(-kt)[/tex]
where,
A = Concentration after time 't'
A₀ = Initial concentration = 0.052 M
k = rate constant = [tex]0.271 s^{-1}[/tex]
t = time = 5.16 s
On substituting the values:
[tex]A=A_0*exp(-kt)\\\\A=0.52*exp(0.271*5.16)\\\\A=0.0128M[/tex]
Thus, the concentration that will remain after will be 0.0128M.
Find more information about Half life here:
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