Biology How Do Stoichiometric Coefficients Affect Reaction Rates?

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Stoichiometric coefficients significantly influence reaction rates by determining how the rate of change in concentration of reactants and products is calculated. For the reaction where 2C forms 2E and 1F, the rate of formation of E is given as 1.6×10^(-4) mol L^(-1) s^(-1), leading to a corresponding rate of usage for reactant C also at 1.6×10^(-4) mol L^(-1) s^(-1). The reaction rate is derived from the stoichiometric coefficients, resulting in a value of 0.8×10^(-4) mol L^(-1) s^(-1) for the overall reaction rate. This calculation reflects that the rate of change for any species is scaled by its stoichiometric coefficient. Understanding this relationship is crucial for accurately determining reaction rates in chemical kinetics.
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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]
 
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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