Question

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).

Answer 1

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

Step-by-step 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^{- (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 ^{-(Q)/(RT_1) }`

For the second rate constant

`k_2 = Ce ^{-(Q)/(RT_2) }`

Now the ratio between the two given rate constant is

`(k_1 )/(k_2) = e^{((Q)/(R) [(1)/((T_2 - 1)/(T_1) ) ] )}`

=>
`ln [(k_1)/(k_2) ] = (Q)/(R) * [(1)/((T_2 -1)/(T_1) ) ]`

substituting values

`ln [(1.05 *10^(-8))/(2.95 *10^(-4)) ] = (Q)/(8.314) * [(1)/((673 -1)/(573) ) ]`

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