Also what does the factor P actually do and what does a catalyst change?

Arrhenius: my Ea for 'rate doubles per 10 K' is 53.6 kJ but the book says 52.9. Who is right?

Two numbers, both for the same textbook statement: 'the rate of a reaction roughly doubles for every 10 °C rise'. I tried to work out $E_a$ for it.

Using

$\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

with $k_2/k_1 = 2$ , $T_1 = 300$ K and $T_2 = 310$ K, I got $E_a \approx 53.6$ kJ/mol. NCERT says $52.9$ kJ/mol for the same kind of problem. Did I do something wrong, or is the book wrong?

Then collision theory. The rate constant in collision theory is $k = P Z_{AB}e^{-E_a/RT}$ . I understand $Z_{AB}$ is the collision frequency and $e^{-E_a/RT}$ is the fraction of collisions with enough energy. But what is $P$ (the 'steric factor' or 'probability factor')? Do I need it for NEET?

Last thing. A friend's notes say 'a catalyst lowers $\Delta H$ so the reaction releases more heat and gives more products'. My gut says that's incorrect. Amma would say I'm anxious for no reason, but I want to be sure about this one before my test.

6 answers & replies

[+]Aman YadavDropperSample· 3d ago

Nobody is wrong. Check the temperatures. NCERT uses 298 K to 308 K. Compute:

$E_a = 2.303\times8.314\times\log 2\times\frac{298\times308}{10} = 5.763\times9178 \approx 52.9\ \text{kJ/mol}$

Yours: 300 and 310: $5.763\times\frac{300\times310}{10} = 5.763\times9300 \approx 53.6$ kJ/mol. Same method, different temperatures. The point is that $E_a$ around 50 kJ/mol gives a doubling near room temperature. It's not an exact law, just a rule of thumb.

[+][–]Kavya NaikDropperSample· 3d agoAuthor
ayyo it was just the temperatures! Both fine. And amma, if you are reading this, yes I will sleep now. Thanks everyone 💛

[+]Rohan Sharma11thSample· 3d ago

catalyst lowers ΔH so more heat is released and equilibrium shifts right, more products. your friend is right imo, catalysts "help" the reaction 💪 that's what we use in factories also, to get more of the product, otherwise why would anybody pay for them. also my uncle works in a plant so I've heard this many times from him

[+][–]Vivek Mishra12thSample· 3d ago
Not right, Rohan. A catalyst provides a different path with a lower activation energy. It does not change the energies of the reactants and products, so $\Delta H$ stays the same, and $\Delta G$ and the equilibrium constant do not change either. It lowers $E_a$ for both the forward and the reverse reaction by the same amount, so both speed up equally, and the system reaches equilibrium faster. It does not give more product at equilibrium. That's why Kavya's gut is correct.

[+]Nisha Verma12thSample· 3d ago

The factor $P$ : not every collision with enough energy leads to a reaction, because the molecules must also hit with the correct orientation. Take $\text{NO} + \text{O}_3 \to \text{NO}_2 + \text{O}_2$ : O₃ has to be hit at the right atom end. $P$ is a number between 0 and 1 (typically much less than 1) that corrects for this. In NCERT-level questions you need the idea in words: effective collisions need (i) energy $\ge E_a$ (threshold energy) and (ii) proper orientation. The formula with $P$ is good to know but numericals are on Arrhenius.

[+]Tushar PawarDropperSample· 3d ago

Handy plot: take logs of the Arrhenius equation,

$\ln k = \ln A - \frac{E_a}{R}\cdot\frac1T$

So a graph of $\ln k$ against $1/T$ is a straight line with slope $-E_a/R$ and intercept $\ln A$ . If the problem gives you the slope (say $-6000$ K), then $E_a = 6000 \times 8.314 \approx 49.9$ kJ/mol. With $\log_{10}k$ instead of $\ln k$ the slope is $-E_a/2.303R$ . Watch for which log is used. Also $e^{-E_a/RT}$ is the fraction of molecules with $E \ge E_a$ , so raising $T$ raises that fraction a lot, which is why rates grow so fast with temperature.

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