# How Do Stoichiometric Coefficients Affect Reaction Rates?

• Biology
• Master1022
In summary, the reaction rate is ##0.8\times10^{–4} mol L^{–1} s^{–1}## and the rate of usage of reactant C is also ##1.6\times10^{–4} mol L^{–1} s^{–1}##.
Master1022
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?
Relevant Equations
kinetics equations
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? "

(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

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]

Master1022
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##.

Chestermiller

## What is kinetics?

Kinetics is the study of the rates of chemical reactions, including the factors that affect the speed of a reaction.

## How do you measure the rate of a reaction?

The rate of a reaction can be measured by monitoring the change in concentration of a reactant or product over time. This can be done using various techniques such as spectrophotometry, titration, or gas chromatography.

## What factors affect the rate of a reaction?

The rate of a reaction can be affected by factors such as temperature, concentration of reactants, surface area, presence of a catalyst, and the nature of the reactants and products.

## What is a rate law?

A rate law is an equation that relates the rate of a reaction to the concentrations of the reactants. It is determined experimentally and can help predict the rate of a reaction under different conditions.

## Why is understanding kinetics important?

Understanding kinetics is important in various fields such as pharmaceuticals, environmental science, and materials science. It allows us to optimize reaction conditions, design more efficient processes, and develop new products.

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