## Quiz 1, Winter 2014 Solving for Activation Energy

Arrhenius Equation: $\ln k = - \frac{E_{a}}{RT} + \ln A$

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### Quiz 1, Winter 2014 Solving for Activation Energy

For the reaction, HO + H2 --> H2O + H, a plot of lnk versus 1/T gives a straight line with a slope equal to -5.1x10^3 K
What is the activation energy for this reaction?

from the Arrhenius EQ, I found that Ea= lnK- lnA/ -RT

However, I don't know how to find the values of lnK, lnA, and T?

I realize that lnA is the intercept to the slope of -5.1 x 10^3, but I'm definitely very stuck.

Thanks for the help!

Kayla Denton 1A
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### Re: Quiz 1, Winter 2014 Solving for Activation Energy

Put the equation into y = mx + b form where y is the dependent variable (lnk), x is the independent variable (1/T), b is the y-intercept and m is the slope.
You get lnk = (-Ea/R)(1/T) + lnA (I wrote 1/T separately rather than combining it into -Ea/RT to make it easier to visualize).
From here you can see that the slope, m, is -Ea/R.
Slope = -Ea/R
Slope x R = -Ea
-Slope x R = Ea

That's how you solve for Ea :)

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### Re: Quiz 1, Winter 2014 Solving for Activation Energy

Thank you so much!

Andre Tsivis 1K
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### Re: Quiz 1, Winter 2014 Solving for Activation Energy

Does R stand for Rate or Gas constant here?

Chem_Mod
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### Re: Quiz 1, Winter 2014 Solving for Activation Energy

Andre Tsivis 1K wrote:Does R stand for Rate or Gas constant here?

It is gas constant.

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