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what would be the initial rate for an experiment where [no]0 5 6.21 3 1018 molecules/cm3 and [o2]0 5 7.36 3 1018 molecules/cm3?

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Final answer:

Initial rates require a change in concentration over time; provided concentrations alone are insufficient without additional data such as reaction stoichiometry or concentration changes.

Step-by-step explanation:

The student has asked for the calculation of an initial rate for a given chemical reaction based on provided concentrations and time intervals. To calculate the initial rate, one must use the change in concentration over the change in time for one of the reactants or products. However, the initial rate can't be determined from the concentrations given
([NO]0= 7.36 × 1018 molecules/cm3) without additional information such as the reaction stoichiometry and change in concentration over a specific period. Typically, initial rates are calculated using concentration changes like Δ[O2]/Δt, and the information provided describes concentrations at a certain point, but not how they change over time.

User Toni Michel Caubet
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The initial rate is:


\[ \text{Initial rate} = (\Delta[A])/(\Delta t) = (\Delta[B])/(\Delta t)\]

To calculate the initial rate of a reaction, you need the change in concentration of a reactant or product over a short period of time (Δ[A] or Δ[B]) and the corresponding change in time (Δt) at the very beginning of the reaction. The initial rate can be represented as:


\[ \text{Initial rate} = (\Delta[A])/(\Delta t) = (\Delta[B])/(\Delta t)\]

However, without additional information about the reaction and the rate expression, it's not possible to calculate the initial rate from just the initial concentrations of the reactants ([NO]0 and [O2]0).

The initial rate depends on the reaction order with respect to each reactant and the rate constant (k) for the specific reaction. The rate expression could be first-order, second-order, etc., and the rate law for the reaction would specify how the concentrations of the reactants affect the rate.

User Peshraw Hasan
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