Physics: The rate at which a metal alloy oxidizes in an oxygen-containing atmosphere is a typical example of the practical utility of the Arrhenius equation. For example, the rate of oxidation of a magnesium alloy is represented by a rate constant, k. The value of k at 300°C is 1.05 * 10-8kg/(m4 # s). At 400°C, the value of k rises to 2.95 * 10-4kg/(m4 # s). Calculate the activation energy, Q, for this oxidation process (in units of kJ/mol).

A. Pascal s principle is the answer....

The rate at which a metal alloy oxidizes in an - id6689818, sofiaisabelaguozdpez

Nopernope11

28.01.2022

The activation energy is $Q = 328.31 \ K J/mol$

Explanation:

From the question we are told that

The rate constant is k

at the temperature $T_1 = 300 = 300 + 273 = 573 \ K$

The value of k is $k_1 = 1.05 *10^{-8} \ kg /m^4 \cdot s$

at temperature $T_2 = 400 ^oC = 400 + 273 = 673 \ K$

The value of k is $k_2 = 2.95 *10^{-4} \ kg /m^4 \cdot s$

The rate constant is mathematically represented as

$k = Ce^{- \frac{Q}{RT} }$

Where Q is the activation energy

R is the ideal gas constant with a value of $R = 8.314 \ J /mol \cdot K$

C is a constant

T is the temperature

For the first rate constant

$k_1 = Ce ^{-\frac{Q}{RT_1} }$

For the second rate constant

$k_2 = Ce ^{-\frac{Q}{RT_2} }$

Now the ratio between the two given rate constant is

$\frac{k_1 }{k_2} = e^{(\frac{Q}{R} [\frac{1}{\frac{T_2 - 1}{T_1} } ] )}$

=> $ln [\frac{k_1}{k_2} ] = \frac{Q}{R} * [\frac{1}{\frac{T_2 -1}{T_1} } ]$

substituting values

$ln [\frac{1.05 *10^{-8}}{2.95 *10^{-4}} ] = \frac{Q}{8.314} * [\frac{1}{\frac{673 -1}{573} } ]$

=> $Q = 328.31 \ K J/mol$

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