- #1
adf89812
- 37
- 1
- TL;DR
- When to Multiply by Constants and do exponentiation in rate law equation ?
When to use exponents or multiply by constants in the rate law?
Click For Summary
SUMMARY
The discussion centers on the application of stoichiometric coefficients in rate laws, specifically how they influence the concentration terms in chemical kinetics. It is established that the concentration of reactants is raised to the power of their stoichiometric coefficients, while the internal structure of the molecules is irrelevant for single-step kinetics. The rate of reaction is expressed as ##k_3[H_2][I]^2##, indicating that a stoichiometric coefficient of two not only affects the exponent but also the rate at which reactants are consumed. The conversation emphasizes that while it is possible to calculate reaction rates without directly doubling them, the stoichiometric coefficients must be incorporated at some stage in the analysis.
PREREQUISITES- Understanding of chemical kinetics and rate laws
- Familiarity with stoichiometric coefficients in chemical reactions
- Knowledge of reaction quotients and their definitions
- Basic grasp of molecular structure and its implications in kinetics
- Study the derivation of rate laws from stoichiometric coefficients
- Explore the concept of reaction mechanisms and their impact on kinetics
- Learn about the role of reaction quotients in equilibrium and kinetics
- Investigate advanced topics in chemical kinetics, such as multi-step reactions
Chemistry students, researchers in chemical kinetics, and professionals involved in reaction mechanism analysis will benefit from this discussion.
- #2
Borek
Mentor
- 29,170
- 4,598
You raise concentration to the power given by the stoichiometric coefficient, internal structure of the molecule (amount of atoms of different elements) doesn't matter (well, it can matter for the mechanism, but for a single step kinetics these are just black boxes doing some magic).
This is not different from the way reaction quotient is defined.
This is not different from the way reaction quotient is defined.
- #3
adf89812
- 37
- 1
where does the doubling in theBorek said:
- #4
Borek
Mentor
- 29,170
- 4,598
Rate of the reaction is ##k_3[H_2][ I]^2## but because of the stoichiometry I is consumed twice as fast.
- #5
adf89812
- 37
- 1
The stoichiometry coefficient of two means both double the speed and square the exponent on the concentration term always?Borek said:
- #6
adf89812
- 37
- 1
the stoichiometry coefficient on the reactant side of the decomposition has two effects always?adf89812 said:
- #7
Borek
Mentor
- 29,170
- 4,598
Many ways to skin that cat. In general it is perfectly possible to do the calculations without doubling the rate, but this stoichiometry coefficient needs to be used at some point.
- #8
adf89812
- 37
- 1
If there were homonuclear triatomic molecules, would that mean cubed the corresponding concentration but don't multiple the rate by three unless the stoichiometry means it's concentration is three times that of the others?Borek said:
- #9
Borek
Mentor
- 29,170
- 4,598
adf89812 said:
Why? I already told you whatever is inside the molecule doesn't matter.
Similar threads
- · Replies 2 ·
- · Replies 8 ·
- · Replies 1 ·
- · Replies 7 ·
- · Replies 8 ·
- · Replies 4 ·
- · Replies 2 ·