I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
Report (any type)/Brief report
22/09/2021
Here’s how you could do that.
The Arrhenius equation does not include a ##T_2##, it only includes a ##T##, the absolute temperature at which the reaction is taking place.
However, you can use the Arrhenius equation to determine a ##T_2##, provided that you know a ##T_1## and the rate constants that correspond to these temperatures.
The Arrhenius equation looks like this
##color(blue)(|bar(ul(color(white)(a/a)k = A * “exp”(-E_a/(RT))color(white)(a/a)|)))” “##, where
##k## – the rate constant for a given reaction
##A## – the pre-exponential factor, specific to a given reaction
##E_a## – the activation energy of the reaction
##R## – the universal gas constant, useful here as ##8.314″J mol”^(-1)”K”^(-1)##
##T## – the absolute temperature at which the reaction takes place
So, let’s say that you know the activation energy of a chemical reaction you’re studying.
You perform the reaction at an initial temperature ##T_1## and measure a rate constant ##k_1##. Now let’s say that you’re interested in determining the temperature at which the rate constant changes to ##k_2##.
You can use Arrhenius equation to write
##k_1 = A * “exp”(-E_a/(R * T_1))##
and
##k_2 = A * “exp”(-E_a/(R * T_2))##
To find ##T_2##, divide these two equations
##k_1/k_2 = color(red)(cancel(color(black)(A)))/color(red)(cancel(color(black)(A))) * (“exp”(-E_a/(R * T_1)))/(“exp”(-E_a/(R * T_2)))##
Use the property of exponents
##color(purple)(|bar(ul(color(white)(a/a)color(black)(x^a/x^b = x^((a-b)), AA x !=0)color(white)(a/a)|)))##
to rewrite the resulting equation as
##k_1/k_2 = “exp”[E_a/R * (1/T_2 – 1/T_1)]##
Next, take the natural log of both sides of the equation
##ln(k_1/k_2) = ln(“exp”[E_a/R * (1/T_2 – 1/T_1)])##
This will be equivalent to
##ln(k_1/k_2) = E_a/R * (1/T_2 – 1/T_1)##
Finally, do some algebraic manipulation to isolate ##T_2## on one side of the equation
##ln(k_1/k_2) = E_a/R * 1/T_2 – E_a/R * 1/T_1##
##1/T_2 = ln(k_1/k_2) + E_a/R * 1/T_1##
Therefore,
##color(green)(|bar(ul(color(white)(a/a)color(black)(T_2 = (R * T_1)/(R * T_1 * ln(k_1/k_2) + E_a))color(white)(a/a)|)))##
I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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