Question

Dinitrogen pentoxide (N2O5) decomposes as follows to nitrogen dioxide and nitrogen trioxide: N,o,()NO,(g)-No, (g) Calculate the average rate of the reaction between consecutive measurement times in the following table. Time (s) IN2O51 (molecules/cm2) 0.00 1.605x1012 1.25 1.529x1012 1.471x1012 2.50 3.75 1.425x1012 5,00 1.389x1012 1OF 14 OUreTIONE Tropospheric ozone is rapidly consumed in many reactions, including (o-(a)ON(a)oN- ()0 Use the following data to calculate the instantaneous rate of the preceding reaction at t 0.000s and t-0.0520 s Express your answers to three significant figures Time (s) INO (M) 0.000 2.00-10 1.80-10 0.0110 0.0270 160-10 0.0520 140-108 0.1020 1.20-10

Answer 1

1)

Rate 1 = -{[N₂O₅]final – [N₂O₅]initial}/Time

= -(1.529 - 1.605) x 10¹²/1.25

= -(- 0.076) x 10¹² / 1.25

= 6.08 x 10¹⁰ molecules/cm³s

Rate 2 = -{[N₂O₅]final – [N₂O₅]initial}/Time

= -(1.471 - 1.529) x 10¹²/1.25

= - (-0.058)x10¹² /1.25

=  4.64 x 10¹⁰ molecules/cm³s

Rate 3 = -{[N₂O₅]final – [N₂O₅]initial}/Time

= -(1.323-1.368) x 10¹²/1.25

= -(-0.045 x 10¹² / 1.25

= 3.60 x 10¹⁰ molecules/cm³s

Rate 4 = -{[N₂O₅]final – [N₂O₅]initial} / Time

= -(1.389 - 1.425) x 10¹²/1.25

  = -(-0.036 x 10¹² / 1.25 )

= 2.88 x 10¹⁰ molecules/cm³s

The average rate of the reaction is = ( Rate 1 + Rate 2 + Rate 3 + Rate 4 ) / 4

= ( 6.08 + 4.64 + 3.60 + 2.88 ) x 10 ¹⁰ / 4

= ( 1.72 x 10¹¹ ) / 4

= 4.3 x 10¹⁰  molecules/cm³s