Respuesta :
Answer:
3.91 minutes
Explanation:
Given that:
Biacetyl breakdown with a half life of 9.0 min after undergoing first-order reaction;
As we known that the half-life for first order is:
[tex]t__{1/2}}= \frac{0.693}{k}[/tex]
where;
k = constant
The formula can be re-written as:
[tex]k = \frac{0.693}{t__{1/2}}[/tex]
[tex]k = \frac{0.693}{9.0 min}[/tex]
[tex]k = 0.077 min^{-1}[/tex]
Let the initial amount of butter flavor in the food be [tex](N_0)[/tex] = 100%
Also, the amount of butter flavor retained at 200°C [tex](N_t)[/tex]= 74%
The rate constant [tex]k = 0.077 min^{-1}[/tex]
To determine how long can the food be heated at this temperature and retain 74% of its buttery flavor; we use the formula:
[tex]\frac{N_t}{N_0}= -kt[/tex]
[tex]t = - (\frac{1}{k}*In\frac{N_t}{N_0} )[/tex]
Substituting our values; we have:
[tex]t = - (\frac{1}{0.077}*In\frac{74}{100} )[/tex]
t = 3.91 minutes
∵ The time needed for the food to be heated at this temperature and retain 74% of its buttery flavor is 3.91 minutes
3.91 minutes is the time the food be heated at this temperature and retain 74% of its buttery flavor.
What is half life time of a reaction?
Half life time is the required time when the concentration of reactant left with the half of the initial concentration.
Given half life time for the first order reaction of the breakdown of Biacetyl = 9 min.
We know that half life time for 1st order reaction is written as:
t = 0.693/k, where
k = rate constant
k = 0.693/9 = 0.077 min⁻¹
Now, required time for the food be heated at this temperature can be calculated by using the rate equation of first order reaction as:
T = -(1/k×ln[N/N₀]), where
N = initial amount of butter flavor in food = 100% (given)
N₀ = amount of butter flavor retained food = 74% (given)
Now putting all these values in the above equation, we get
T = - (1/0.077 × ln[74/100])
T = 3.91 minutes
Hence, 3.91 minutes is the time the food be heated at this temperature.
To know more about half life time, visit the below link:
https://brainly.com/question/14936355
