# For some transformation having kinetics that obey the Avrami equation, the parameter n is known...

## Question:

For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 1.2. If after 119 s, the reaction is 50% complete how long will it take the transformation to go to 87% completion?

## Avrami's equation:

The mathematical relation that describes the information for the transition of solid substances from one phase to another phase is called Avrami's equation. Avrami's equation is used to analyze, as well as to interpret various chemical reactions and laboratory inspections.

## Answer and Explanation:

**Given data:**

- The time is, {eq}t = 119\;{\rm{s}} {/eq}.

- The Avrami's exponential constant is, {eq}n = 1.2 {/eq}.

- The initial transformation rate is, {eq}{Y_1} = 50\% {/eq}.

- The final transformation rate is, {eq}{Y_2} = 87\% {/eq}.

The Avrami's equation is given as,

{eq}1 - {Y_1} = {e^{ - K{t^n}}} {/eq}

Here, **K** is another exponential constant.

Solve by substituting the values as,

{eq}\begin{align*} 1 - {Y_1} = {e^{ - K{t^n}}}\\ 1 - 0.5 = {e^{ - K \times {{119}^{1.2}}}}\\ 0.5 = {e^{ - K \times 309.4969}} \end{align*} {/eq}

Taking natural log on both sides of the above equation as,

{eq}\begin{align*} 0.5 &= {e^{ - K \times 309.4969}}\\ \ln \left( {0.5} \right) = \ln \left( {{e^{ - K \times 309.4969}}} \right)\\ K &= 2.2395 \times {10^{ - 3}} \end{align*} {/eq}

Taking Avrami?s equation for final transformation rate as,

{eq}\begin{align*} 1 - {Y_2} &= {e^{ - K{t^n}}}\\ 1 - 0.87 &= {e^{ - 2.2395 \times {{10}^{ - 3}} \times {t^{1.2}}}}\\ 0.13 &= {e^{ - 2.2395 \times {{10}^{ - 3}} \times {t^{1.2}}}} \end{align*} {/eq}

Taking natural log both sides in above equation as,

{eq}\begin{align*} 0.13 &= {e^{ - 2.2395 \times {{10}^{ - 3}} \times {t^{1.2}}}}\\ \ln \left( {0.13} \right) &= \ln \left( {{e^{ - 2.2395 \times {{10}^{ - 3}} \times {t^{1.2}}}}} \right)\\ - 2.04022 &= - 2.2395 \times {10^{ - 3}} \times {t^{1.2}}\\ t &= 293.92\;{\rm{s}} \end{align*} {/eq}

Therefore, for **87%** completion the required time is **293.92 seconds**.

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