Solutions to Additional Problems  
For Chapter 13

Problem 1

The compound, nitrogen dioxide decomposes to nitrogen monoxide and diatomic oxygen, according to the reaction:

2 NO₂(g) ® 2 NO(g) + O₂(g).  The concentration versus time data below were obtained for this reaction at 330^oC.  The initial concentration was 0.0100 mol/L.

  a)       What is the order of the reaction, and what is the rate law?

  b)       Write the concentration-time relationship.

  c)       From the appropriate graph, estimate the rate constant.  Include the units.

  d)        During the course of the experiment, the laboratory clock stopped working.  The concentration at the end of the experiment was found to be 0.0015 mol/L.  How much time had elapsed since the beginning of experiment?

  e)       What concentration of NO₂ will be left after 1.00 min?  

a) Second order because a plot of 1/(NO₂⁾ vs time  is linear.  Rate = -D(NO₂)/(2Dt) = k (NO₂)²

b) 1/(NO₂⁾ - 1/(NO₂⁾_o = kt

c) k = slope = (330 - 100) L/mol/(300 - 0)s = 0.767 L/(mol-s)

d) 1/(0.0015 M) - 1/(0.01M) = 0.767 L/(mol-s) t

t = 739 s

e)  1/(NO₂) = 1/(0.010 M) +0.767L/(mol-s)(60s) = 146.0 L/mol

(NO₂) = 0.00685 M

 Problem 2

The hydrolysis of sucrose, in which a sucrose molecule is broken down into a glucose molecule and a fructose molecule, is a part of the digestive process.  How strongly does the rate depend on the body temperature?  Calculate the rate constant for the hydrolysis of sucrose at 35^oC, given that k = 1.0 x 10^-3 M^-1 s^-1 at 37^oC (normal body temperature), and the activation energy of the reaction is 108 kJ mol^-1.  Ans.  7.6 x 10^-4 M^-1s^-1  
ln(k₂/k₁) = E_a/R(1/T₁-1/T₂)
k₂ = k at 35^oC
k₁ = k 37^oC = 1 x 10^-3 M^-1s^-1
E_a = 108 x 10³ J mol^-1
R = 8.314 J K^-1 mol^-1
T₁ = 273 + 37 = 310 K
T₂ = 273 + 35 = 308 K
ln(k₂/k₁) =108 x 10³ J mol^-1/(8.314 J K^-1 mol^-1 )(1/310 K - 1/308 K)
ln(k₂/k₁) = -0.2721
k₂/k₁ = e^-0.2721 = 0.762
k₂ = k₁ x 0.762 = 1 x 10^-3 M^-1s^-1 x 0.762 = 7.6 x10^-4 M^-1s^-1