Answer:
D) 2.3 x 10⁻¹ s⁻¹
Explanation:
The rate constant is related to the activation energy through the formula:
k= Ae^(-Eₐ /RT)
where A is the collision factor, Eₐ the activation energy, R is the gas constant ( 8.314 J/Kmol ) , and T is the temperature (K)
So a plot of lnk versus 1/T ( Arrehenius plot ) gives us a straight line with slope equal -Eₐ/R and intercept lnA
lnk = -(Eₐ/T)(1/T) + lnA
which has the form y= mx + b
In this problem, we can use the data provided to:
a) Using a calculator determine the slope and intercept and then calculate the value of rate constant at 320 ºC, or
b) Plot the data and determine the equation of the best line , and answer the question for k @ 320 ºC by reading the value from the plot.
Once you do the plot, the resulting equation is:
y = - 19 x 10³ x + 30,582 ( R² = 0.999 )
So for T = 320 + 273 K = 593 K
Y = 19 x 10³ X + 30.58
So for T = (320 + 273)K = 593 K
Y = -19 x 10³ ( 1/593) + 30.58 = -32.04 +30.58 = - 1.46
and then since
y = lnk ⇒ e^y = k
k= e^-1.46 = 2.3 x 10⁻¹ s⁻¹
Note: there is an error of transcription in the value for T = 472.1 ( 1/T = 2.118 x 10⁻³ and not 2.228 x 10⁻³). You can recognize this mistake if you plot the data and notice it produces an outlier.
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