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3 years ago

6

For some transformation having kinetics that obey the Avrami equation , the parameter n is known to have a value of 1.1. If, aft

er 114 s, the reaction is 50% complete, how long (total time) will it take the transformation to go to 87% completion
y = 1 - exp(-kt^n)

1 answer:

ivanzaharov [21]3 years ago

7 0

Answer:

total time = 304.21 s

Explanation:

given data

y = 50% = 0.5

n = 1.1

t = 114 s

y = 1 - exp(-kt^n)

solution

first we get here k value by given equation

y = 1 - $e^{(-kt^n)}$ ...........1

put here value and we get

0.5 = 1 - e^{(-k(114)^{1.1})}

solve it we get

k = 0.003786 = 37.86 × $10^{4}$

so here

y = 1 - $e^{(-kt^n)}$

1 - y = $e^{(-kt^n)}$

take ln both side

ln(1-y) = -k × $t^n$

so

t = $\sqrt[n]{-\frac{ln(1-y)}{k}}$ .............2

now we will put the value of y = 87% in equation with k and find out t

t = $\sqrt[1.1]{-\frac{ln(1-0.87)}{37.86*10^{-4}}}$

total time = 304.21 s

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