Calculating Rate Constant for 2N2O5 Reaction

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SUMMARY

The calculation of the rate constant for the first-order reaction 2N2O5 → 4NO2 + O2 is determined using the half-life formula. Given a half-life (t1/2) of 153.0 seconds, the rate constant (k) is calculated using the equation k = ln(2) / t1/2. This results in a rate constant of approximately 0.0045 1/sec. The discussion confirms the relationship between half-life and the rate constant for first-order reactions.

PREREQUISITES
  • Understanding of first-order reaction kinetics
  • Familiarity with the half-life concept in chemical reactions
  • Knowledge of natural logarithms and their application in kinetics
  • Basic proficiency in differential equations related to reaction rates
NEXT STEPS
  • Study the derivation of the first-order rate law and its applications
  • Learn about the Arrhenius equation and its impact on rate constants
  • Explore the concept of reaction mechanisms and their influence on rate laws
  • Investigate the effects of temperature on reaction rates and constants
USEFUL FOR

Chemistry students, chemical engineers, and researchers in reaction kinetics will benefit from this discussion, particularly those focused on understanding and calculating rate constants for chemical reactions.

parwana
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At a certain temperature the reaction

2N2O5 --> 4NO2 + O2

is a first order reaction with t1/2 = 153.0 sec

What is the rate constant for this reaction (1/sec) ?

Enter a numeric answer only, do not include units in your answer


Ok I know the equation is t1/2= ln2/k

so isn't k the rate constant, and wouldn't the answer be ln2/153 sec= .0045??
 
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The first order equation is -

\frac{d [A]}{dt} = - k [A]

where k is the rate constant.

The solution, as you correctly mentioned, is -

ln \frac{[A]}{[A_o]} = - k t, where [A] is the concentration at time t, and Ao is the initial concentration, at t=0.

The definition of half-life is-

ln \frac{[A_o]/2}{[A_o]} = - k t_{1/2}, or

ln 2 = k t1/2
 

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