science

For the reaction: 2 N2O5 (g) → 4 NO2 (g) + O2 (g)

The experimental data for this reaction is :

t = 0.0 s [N2O5] = 0.80 M [NO2] = 0.0 M [O2] = 0.0 M

t = 20.0 s [N2O5] = 0.60 M [NO2] = 0.40 M [O2] = 0.10 M

What is the average rate of reaction during this time interval?

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩
  1. avg rate = #M/time
    That will be in M/s, right?
    And I think you're playing pretty fast and loose with those = signs

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  2. The average rate of this reaction is -1/2[delta(N2O5)/delta t] = -1/2[(0.60M - 0.80M)/(20s-0s)] = -1/2(-.20/20) = -1/2(-0.01) = + 0.005 M/s

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩

Respond to this Question

First Name

Your Response

Similar Questions

  1. Chemistry

    The mechanism for the reaction described by 2N2O5(g) ---> 4NO2(g) + O2(g) is suggested to be (1) N2O5(g) (k1)--->(K-1) NO2(g) + NO3(g) (2) NO2(g) + NO3(g) --->(K2) NO2(g) + O2(g) + NO(g) (3) NO(g) + N2O5(g) --->(K3) 3NO2(g)

  2. Chemistry

    The following reaction mechanism has been proposed for a reaction: (slow) NO2 + NO2-> NO3 + NO (fast) NO3 + CO -> CO2 + NO2 A. Write the equation for the overall reaction from the mechanism above. B.Write the rate law for the

  3. AP chemistry

    The mechanism for the reaction of nitrogen dioxide with carbon monoxide to form nitric oxide and carbon dioxide is thought to be NO2+NO2 > NO3+NO slow NO3+CO > NO2+CO2 fast Write the rate law expected for this mechanism. What is

  4. chemistry

    Calculate ∆H0 for the reaction 2 N2(g) + 5 O2(g) −→ 2 N2O5(g) given the data H2(g) + 1 2 O2(g) −→ H2O(ℓ) ∆H0 f = −290 kJ/mol N2O5(g) + H2O(ℓ) −→ 2 HNO3(ℓ) ∆H0 = −77.8 kJ/mol 1 2 N2(g) + 3 2 O2(g) + 1 2

  1. chem

    For the reaction: 2 N2O5 (g) → 4 NO2 (g) + O2 (g) Later during the reaction, the rate of consumption of N2O5 was determined to be 0.0080 M/s. What was the rate of formation of NO2 and of O2 at this point in time?

  2. Chemistry

    O3 + NO --> O2 + NO2 (all in gas state) Calculate the change in enthalpy for the reaction at room temp. using the following data ^Hf: O3 = 143 NO = 90 NO2 = 33 So, I have 143+90--> X + 33. I don't know what the enthalpy of O2 is.

  3. Chemistry

    Dinitrogen pentoxide, N2O5, decomposes by first-order kinetics with a rate constant of 0.15 s-1 at 353 K. (a) What is the half-life (in seconds) for the decomposition of N2O5 at 353 K? (b) If [N2O5]0 = 0.0579 mol·L-1, what will

  4. Chem

    Calculate delta H for the formation of one mole of N2O5 from the elements at 25 degrees C using the following data. 2H2 + O2 - 2H2O delta H = -571.6 kJ N2O5 + H2O - 2HNO3 deltaH = -73.7 kJ 1/2N2 + 3/2O2 + 1/2H2 - HNO3 delta H =

  1. chemistry

    For the reaction: 2 N2O5 (g) → 4 NO2 (g) + O2 (g) The experimental data for this reaction is : t = 0.0 s [N2O5] = 0.80 M [NO2] = 0.0 M [O2] = 0.0 M t = 20.0 s [N2O5] = 0.60 M [NO2] = 0.40 M [O2] = 0.10 M How much N2O5 was

  2. chemistry

    For the decomposition of gaseous dinitrogen pentoxide (shownbelow), 2 N2O5(g).....> 4NO2(g) + O2(g) the rate constant is k = 2.8 10-3 s-1 at 60°C. The initial concentration of N2O5 is 1.52 mol/L. (a) What is [N2O5] after 5.00

  3. Chemistry

    What mass of N2O5 will result from the reaction of 6.0 mol of NO2 if there is a 61.2% yield in the reaction?

  4. Chemistry

    What is the percentage yield of N2O5 when 68.0g of O2 reacts With N2 according to the reaction equation, with 72.6g of N2O5 being obtained? 2N2 + 5O2 -> 2N2O5

View more similar questions or ask a new question.