Question

#1 please

sactions and kinetic 6.50 x 10's INOCHOD. = 3.00 M 1. (30 pts) Kinetics Consider the following reactions and kinetic information: 1 2N,Os(9) - 4NO2(g) + O2(g), 1st order, k = 6.50 x 10's. [ NOsJo = 2.00 M. il CH CHO - CH + CO, 2nd order, k = 6.50 x 10'M's, (CH3CHOJo = 3.00 M Will more dinitrogen pentoxide (1) or more acetaldehyde (ll) remain after 2500 s? 2. 25 pel The Reaction Pendi (28) Sketch reaction, and local shoppe reaction enthalpy and an active complex anon (26) What effect would a catalyst have cont (2c) What would have the mate constant for the reaction in 2b? (20) Write a mathematical energy for the wardro are related in on the show how e and action and con many reverseennon )

Answer 1

2N2O5 ---> 4NO2 +O2

given

K= 6.5 x10^-3 S-1
initial concentration , Ao;; [N2O50] = 2.00 M
time taken = 2500 s
final concentration, A;; [N2O5]= ??
Use the formula for 1st order kinetics

A= Aoe^-kt
A= 2 M e^[-6.5 x10^-3*2500 s]
= 1.753 x10^-7 M
thus final concentration of N2O5 =1.753 x10^-7 M
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CH3CHO +--> CH4 + CO

given
K= 6.5 x10^-3 S-1
initial concentration , Ao = 2.00 M
time taken = 2500 s
final concentration, A= ??
For 2nd order reaction ,integrated form of rate law,

1/A = kt + 1/Ao
=6.5 x10^-3 S-1*2500 s + 1/3

= 16.25 +0.33 =17.0833

A = 1/17.083= 0.05853 M
thus final concentration of CH3CHO = 0.05853 M

Thus after 2500 s ,CH3CHO will remain more than N2O5
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