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:

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.