Zero Order Integrated Rate Law and Concentration Change Calculation

Zero Order Integrated Rate Law

Definition of Zero Order Reaction

  • A zero order reaction is a chemical reaction in which the rate of the reaction is constant and does not depend on the concentration of the reactants.

Zero Order Rate Law

  • The rate of a zero order reaction can be described by the equation: extRate=kext{Rate} = k where:
    • Rate: The speed at which the reaction takes place (change in concentration over time).
    • k: Rate constant (units: M·s^(-1)).

Integrated Rate Law for Zero Order Reactions

  • The integrated rate law for a zero order reaction is given by: [A]=[A]0−kt[A] = [A]_0 - kt where:
    • [A]: Concentration of reactant (ammonia in this case) at time t.
    • [A]_0: Initial concentration of reactant.
    • k: Rate constant.
    • t: Time.

Example Reaction

  • Reaction being analyzed:
    2extNH<em>3(g)→ N</em>2(g)+3extH2(g)2 ext{ NH}<em>3 (g) \rightarrow \text{ N}</em>2 (g) + 3 ext{ H}_2 (g)

  • Given data for the example:

    • Rate constant, k = 0.0042 M·s^(-1)
    • Initial amount of ammonia, (ext[NH<em>3]</em>0=250extmmol=0.250extL( ext{[NH}<em>3]</em>0 = 250 ext{ mmol} = 0.250 ext{ L}

Steps to Solve the Problem

  1. Calculate the Initial Concentration, [NH₃]_0:

    • Volume of the flask = 250 mL = 0.250 L
    • Initial moles of ammonia = 250 mmol = 0.250 mol
    • Initial concentration:
      [NH<em>3]</em>0=0.250extmol0.250extL=1.00extM[NH<em>3]</em>0 = \frac{0.250 ext{ mol}}{0.250 ext{ L}} = 1.00 ext{ M}

  2. Determine the Final Concentration, [NH₃]:

    • Final moles of ammonia = 125 mmol = 0.125 mol
    • Final concentration:
      [NH3]=0.125extmol0.250extL=0.500extM[NH_3] = \frac{0.125 ext{ mol}}{0.250 ext{ L}} = 0.500 ext{ M}

  3. Apply the Integrated Rate Law:

    • Using the integrated rate law, substituting known values:
      [NH<em>3]=[NH</em>3]0−kt[NH<em>3] = [NH</em>3]_0 - kt
    • Substitute [NH3], [NH₃]_0, and k into the equation:
      0.500extM=1.00extM−(0.0042extM⋅s−1)t0.500 ext{ M} = 1.00 ext{ M} - (0.0042 ext{ M·s}^{-1}) t

  4. Solve for Time, t:

    • Rearranging gives:
      (0.0042extM⋅s−1)t=1.00extM−0.500extM(0.0042 ext{ M·s}^{-1}) t = 1.00 ext{ M} - 0.500 ext{ M}
    • Calculate:
      (0.0042extM⋅s−1)t=0.500extM(0.0042 ext{ M·s}^{-1}) t = 0.500 ext{ M}
    • Therefore:
      t=0.500extM0.0042extM⋅s−1≈119.05extst = \frac{0.500 ext{ M}}{0.0042 ext{ M·s}^{-1}} \approx 119.05 ext{ s}

Conclusion

  • The time required for the concentration of ammonia to decrease from 250 mmol to 125 mmol is approximately 119.05 seconds.
  • Ensure to round the answer according to significant digits, in this case:119 s.

Notes

  • It is important to check that all units are consistent and to round the final answer appropriately to maintain significant figures as given in the initial values provided.