0
$\begingroup$

The rate constant for the reaction of hydrogen with iodine is $\pu{2.45E-4 M-1 s-1}$ at 302 °C and $\pu{0.905 M-1 s-1}$ at 508 °C.

a. calculate the activation energy and Arrhenius preexponential factor for this reaction.

b. What is the value of the rate constant at 400 °C ?


I'm a bit confused because I get a negative value for the activation energy $E_a$

We have at 302 °C and 508 °C respectively $$k_1 = \pu{2.45E-4 M-1 s-1}$$ $$k_2 = \pu{0.905 M-1 s-1}$$

We convert in Kelvin: $$T_1 = 302 °C = 575.15 K $$ $$T_2 = 508 °C = 781.15 $$

The Arrhenius Equation is $$\ln{k} = \ln{A} - \frac{E_a}{RT}$$

So we have $$\ln{k_1} = \ln{A} - \frac{E_a}{RT_1}$$ $$\ln{k_2} = \ln{A} - \frac{E_a}{RT_2}$$

We want to solve for $E_a$. We subtract the second equation from the first: $$\ln{k_1} - \ln{k_2} = - \frac{E_a}{R}(\frac{1}{T_1} - \frac{1}{T_2})$$

Solving for $E_a$, we get: $$E_a = - R \times \frac{\ln{k_1} - \ln{k_2}}{\frac{1}{T_1} - \frac{1}{T_2}}$$

With $R = 8.314 J/mol K$, $T_1 = 575.15 K$, $T_2 = 781.15$, $k_1 = \pu{2.45E-4 M-1 s-1}$, $k_2 = \pu{0.905 M-1 s-1}$

(Link to WolframAlpha calculation)

I get $E_a = 1.48948 \times 10^5 J/mol$

Then using the Arrhenius equation with $k_1$ or $k_2$: $$A = k_1 \times \exp{\frac{E_a}{RT_1}}$$

(Link to WolframAlpha calculation)

I get $A = \pu{8.26E9 M-1 s-1}$

But I'm confused, why the Activation Energy is negative ? Or did I miss something ? I checked all my calculations

EDIT: the error was due to a missing negative sign for $E_a = ...$. Now it makes more sense.

$\endgroup$
5
  • $\begingroup$ @Poutnik So how should I formulate the title ? "Why do I get a negative activation energy ?" $\endgroup$
    – wengen
    Dec 1, 2023 at 10:50
  • 1
    $\begingroup$ Instead of $2.45 \times 10^{-4} M^{-1} s^{-1}$ ( $2.45 \times 10^{-4} M^{-1} s^{-1}$ ), try $\pu{2.45E-4 M-1 s-1}$ ($\pu{2.45E-4 M-1 s-1}$). // As a simplified rule, only symbols for variables, physical(chemical) quantities or physical constants are in italic, all the rest is upright. $\endgroup$
    – Poutnik
    Dec 1, 2023 at 11:02
  • $\begingroup$ @Poutnik Thank you very much for your help. The error was in fact due to the missing negative sign. Now it makes more sense. I get $E_a = 148957$ J/mol. This gives an Arrhenius Pre-exponential factor of $\pu{8.26E9 M-1 s-1}$. For b), can we simply use the same formula with $T = 400$ ° C $= 673.15$ K as temperature ? This would give $\pu{8.26E9 \exp{(−148957​/(8.314×673.15))}}$ $= \pu{0.0228 M-1 s-1}$ $\endgroup$
    – wengen
    Dec 1, 2023 at 11:20
  • $\begingroup$ @Poutnik Feel free to post an answer if you want (even just saying that it was due to the missing negative sign). I will accept it. Otherwise I will answer my own question later $\endgroup$
    – wengen
    Dec 1, 2023 at 11:25
  • 1
    $\begingroup$ It was nothing. But consider if "Always doublecheck signs in your calculations if you get weird results" is the good answer material. :-) $\endgroup$
    – Poutnik
    Dec 1, 2023 at 11:30

1 Answer 1

1
$\begingroup$

A mistake must have happened in your calculations, because, when I do them, I obtain :

ln$k_1 = - 8.314$;

ln$k_2 = - 0.100$;

ln$k_1$ - ln$k_2$ = $-8.314 + 0.100 = -8.214$

$\frac{1}{T_1} - \frac{1}{T_2}$ = $ \frac{1}{575} - \frac{1}{781} = 1.76·10^{-3} - 1.97·10^{-3} = - 0.21 ·10^{-3} $ K$^{-1}$

$E/R = \frac{-8.214}{-0.21·10^{-3} \mathrm{K}^{-1}} = + 39 100$ K.

$E = 39100$K$ · 8,314$ $\frac{J}{mol·K} = + 325.2$ kJ/mol

It is a positive result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.