When to use exponents or multiply by constants in the rate law?

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In rate laws, concentrations are raised to the power of their stoichiometric coefficients, while the internal structure of molecules is generally irrelevant for single-step kinetics. The reaction rate is influenced by the stoichiometry, which can affect both the exponent on concentration terms and the overall rate. For example, if a reactant is consumed twice as fast, its concentration is squared in the rate expression. The stoichiometric coefficient can have dual effects on reaction speed and concentration exponent, but calculations can be performed without explicitly doubling the rate. Ultimately, the internal composition of molecules does not impact these calculations significantly.
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When to Multiply by Constants and do exponentiation in rate law equation ?
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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.
 
Borek said:
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.
where does the doubling in the
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come from?
 
Rate of the reaction is ##k_3[H_2][ I]^2## but because of the stoichiometry I is consumed twice as fast.
 
Borek said:
Rate of the reaction is ##k_3[H_2][ I]^2## but because of the stoichiometry I is consumed twice as fast.
The stoichiometry coefficient of two means both double the speed and square the exponent on the concentration term always?
 
adf89812 said:
The stoichiometry coefficient of two means both double the speed and square the exponent on the concentration term always?
the stoichiometry coefficient on the reactant side of the decomposition has two effects always?
 
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.
 
Borek said:
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.
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?
 
adf89812 said:
If there were homonuclear triatomic molecules, would that mean cubed the corresponding concentration

Why? I already told you whatever is inside the molecule doesn't matter.
 
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