0
$\begingroup$

Context: The rate of appearance of $\ce{O2}$ in the reaction: $\ce{2O3 (g) -> 3O2 (g)}$ is $\pu{0.250 M/s}$. So, over the first $\pu{5.50 s}$, will it form $\pu{1.38 M}$, or $\pu{4.13 M}$ of oxygen during the given time?

Which is right? Here are my solutions:

  1. $\pu{0.250 M/s} \cdot \pu{5.50 s} = \pu{1.38 M}$

  2. $\pu{0.250 M/s} \cdot \pu{5.50 s} \cdot 3 = \pu{4.13 M}$, (3 in the calculation is the coefficient of the product in the reaction: $\ce{2 O3 (g) -> \mathbf{3} O2 (g)}$).

I feel like the first one is right? I am overthinking this, and I need to make sure that I am doing this right.

$\endgroup$
5
  • 2
    $\begingroup$ reaction_rate = abs(component_rate)/stoichiometric_coefficient $\endgroup$
    – Poutnik
    Jan 11 at 13:20
  • 1
    $\begingroup$ Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers. $\endgroup$
    – Buttonwood
    Jan 11 at 13:47
  • $\begingroup$ Hello, @Poutnik. Thank you for responding. May I ask what "abs" and "component_rate" refer? I am a bit confused since I have a different formula given by my instructor. $\endgroup$ Jan 12 at 10:27
  • $\begingroup$ Thank you so much, @Buttonwood. I will use mhchem on my next posts. $\endgroup$ Jan 12 at 10:27
  • 2
    $\begingroup$ abs is math function giving an absolute value. Component rate is the time derivative of the component(=reactant or product) concentration ( or partial pressure or total amount, depending on scenario). // Good practice is starting with symbolic algebraic formulas and keeping it this way until all formulas are ready to finally put there literal numbers. It helps focusing on principles and improves orientation, spotting mistakes, the text reusability and its permanent value. $\endgroup$
    – Poutnik
    Jan 12 at 10:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.