How Do Stoichiometric Coefficients Affect Reaction Rates?

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SUMMARY

The discussion focuses on how stoichiometric coefficients influence reaction rates, specifically in the context of a reaction where 2 moles of C react to form 2 moles of E and 1 mole of F. The rate of formation of E is given as 1.6×10–4 mol L–1 s–1. The reaction rate is calculated using the formula r = (1/να) d[Xα]/dt, leading to a reaction rate of 0.8×10–4 mol L–1 s–1 for reactant C, which is half the rate of formation of E due to the stoichiometric coefficients.

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Chemistry students, chemical engineers, and researchers in the field of chemical kinetics will benefit from this discussion, particularly those interested in understanding the relationship between stoichiometric coefficients and reaction rates.

Master1022
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Homework Statement
For the reaction the following stoichiometric coefficients have been determined: 2 ##C## react to form 2 ##E## and 1 ##F##. The rate of formation of E is ##1.6 \times 10^{–4} mol L^{–1} s^{–1}##. What is the reaction rate and rate of usage of reactant C?
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Hi,

I was attempting the following question and don't know how to find the 'reaction rate':
"For the reaction the following stoichiometric coefficients have been determined: 2 ##C## react to form 2 ##E## and 1 ##F##. The rate of formation of E is ##1.6\times10^{–4} mol L^{–1} s^{–1}##. What is the reaction rate and rate of usage of reactant C? "
Screen Shot 2021-03-31 at 8.16.19 AM.png


(note this form of the equation doesn't have the coefficients in front of it)

Attempt:
We can write the equations for the formation of E and C as follows:
\frac{de}{dt} = k_b c^2 - k_{b'} e^2 f = 1.6\times10^{–4} mol L^{–1} s^{–1}
\frac{dc}{dt} = k_{b'} e^2 f - k_b c^2
The second expression is just -1 times the first expression and I am assuming the: rate of formation = -1 * rate of usage. That all leads to the rate of usage of reactant C is ##1.6 \times10^{–4} mol L^{–1} s^{–1}##.

Now I am confused on how to find the reaction rate. The answer says it should be ##0.8\times10^{–4} mol L^{–1} s^{–1}## which I can see is a factor of 2 (or 0.5) away from our current rate, but I don't know how to connect the two. A search on google shows that:
\text{reaction rate} = \frac{-\Delta \text{[reactants]}}{\Delta t}
but it isn't clear how I use that expression to get the required answer.

Thanks for any help
 
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If the stoichiometric coefficient of a certain species ##X_{\alpha}## is ##\nu_{\alpha}## then the rate of reaction is defined to be$$r = \frac{1}{\nu_{\alpha}} \frac{d[X_{\alpha}]}{dt}$$[Note that here the stoichiometric coefficients are defined such that ##\nu_{\alpha} < 0## if ##X_{\alpha}## is a reactant; some texts instead define all of the ##\nu_{\alpha}## to be strictly positive and insert the signs in equations involving them by hand]
 
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Many thanks - that indeed leads to the 0.8! Hadn't seen that before.

etotheipi said:
If the stoichiometric coefficient of a certain species ##X_{\alpha}## is ##\nu_{\alpha}## then the rate of reaction is defined to be$$r = \frac{1}{\nu_{\alpha}} \frac{d[X_{\alpha}]}{dt}$$[Note that here the stoichiometric coefficients are defined such that ##\nu_{\alpha} < 0## if ##X_{\alpha}## is a reactant; some texts instead define all of the ##\nu_{\alpha}## to be strictly positive and insert the signs in equations involving them by hand]
 
No problem! The rate of change of concentration of a particular species is only scaled by its stoichiometric coefficient so that no matter which particular species you choose to measure, you always end up with the same value of ##r##.
 
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