# Reaction kinetics exercise for hydrogen iodide synthesis

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

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

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

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.

• @Poutnik So how should I formulate the title ? "Why do I get a negative activation energy ?" Dec 1, 2023 at 10:50
• 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. Dec 1, 2023 at 11:02
• @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 $\pu{8.26E9 \exp{(−148957​/(8.314×673.15))}}$ $= \pu{0.0228 M-1 s-1}$ Dec 1, 2023 at 11:20
• @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 Dec 1, 2023 at 11:25
• It was nothing. But consider if "Always doublecheck signs in your calculations if you get weird results" is the good answer material. :-) Dec 1, 2023 at 11:30

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.