0
$\begingroup$

When given the reaction: $\ce{NO2}$ reacts with $\ce{O2}$ to produce $\ce{N2O5}$, how do I calculate the rate of consumption of $\ce{NO2}$ when the rate of the reaction= 0.02 M/s?

I am confused about which of these expressions of reaction rate can be used to calculate the rate of consumption of $\ce{NO2}$.

a ) r = 0.02 = (1/4) × rate of consumption of $\ce{NO2}$

Or :

b) r = 0.02 = (1/2) × rate of consumption of $\ce{NO2}$

Does the rate of the reaction changed when we multiply balanced reaction equation by integer? Why?

$\endgroup$
  • $\begingroup$ I think a good starting point would be to write out the balanced equation for NO2 plus O2 to N2O5. $\endgroup$ – airhuff Feb 3 '17 at 22:28
  • $\begingroup$ @airhuff :what happens to the rate of reaction on doubling the stoichiometry of the equation at constant concentrations? $\endgroup$ – Adnan AL-Amleh Feb 4 '17 at 3:14
  • $\begingroup$ To your first comment: your equation has 4 N's on the left and 8 on the right; not balanced. While there are an infinite number of balanced equations, what you want is the lowest factors. Hint: 4NO2 is a good start but the number of O2 and N2O5 molecules are incorrect. This is necessary to know to figure out how much NO2 is consumed per N2O5 produced. $\endgroup$ – airhuff Feb 4 '17 at 3:23
  • $\begingroup$ @airhuff :thank you : We can start with many multiplication of the balanced equation like this :8NO2 +2 O2 -----> 4N2O5 , how to calculate the rate of consumption of NO2?Does it change when doubling the stoichiometry of the equation? $\endgroup$ – Adnan AL-Amleh Feb 4 '17 at 22:01

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.