Rates of Reaction Worksheet Answers
1.	    
(a)	Colour – NO2 (g); pH – HNO3 is an acid
(b)	Formation of a precipitate; conductivity
(c)	pH – HNO3 is an acid; formation of a gas
   
2.	When the molecules collided they did not have sufficient energy.     

3.	    
Large crystals
-	Low surface area
-	Fewer collisions between sugar molecules and O2
-	Slow rate of reaction	Small Crystals
-	High surface area
-	Lots of collisions between sugar molecules and O2
-	Faster rate of reaction	
     
4.	   
r α	Number of collisions x
Concentration
Surface area
temp	Fraction of effective collisions
temp
catalyst
chemical nature	
      
5.	    

   

6.	4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)  

r of NH3 = -1.10 mol NH3 / min

r of O2 = r NH3 x mol ratio = - 1.10 mol NH3    x   5 mol O2 = - 1.38 mol O2 / min
				       min           4 mol NH3

r of NO = - r NH3 x mol ratio = - (- 1.10 mol NH3)    x   4 mol NO = 1.10 mol NO / min
				           min                 4 mol NH3

r of H2O = - r NH3 x mol ratio = -(- 1.10 mol NH3)    x   6 mol H2O =  1.65 mol H2O / min
				             min                4 mol NH3

7.	   
r = k [(C2H5)2(NH)2]m [I2]n   

trial 1		1.08 x 10-4 = k (0.15)m (0.25)n  
trial 2		1.56 x 10-4 = k (0.15)m (0.362)n  
trial 3		2.30 x 10-4 = k (0.20)m (0.40)n  
trial 4		3.44 x 10-4 = k (0.30)m (0.40)n  

r = k  [(C2H5)2(NH)2] [I2]

3.44 x 10-4 = k (0.30) (0.40)
k = 3.8 x 10-3 L/mol h

r = 3.8 x 10-3 L/mol h  [(C2H5)2(NH)2] [I2]

	trial 2		1.56 x 10-4 = k (0.15)m (0.362)n  
trial 1		1.08 x 10-4 = k (0.15)m (0.25)n  

1.46 = (0.362/0.25)m = 1.48m
n = 1

trial 4		3.44 x 10-4 = k (0.30)m (0.40)n
trial 3		2.30 x 10-4 = k (0.20)m (0.40)n  

1.5 = (0.3/0.2)m
m = 1

r = 3.8 x 10-3 L/mol h  [(C2H5)2(NH)2] [I2]
5.00 x 10-4 = k (x) (0.50)
[(C2H5)2(NH)2]  = 1 x 10-3 mol/L

8.	  
r = k  [I‑]m [BrO3-]n [H+]p

trial 1 	8.89 x 10-5 = k (0.002)m (0.008)n (0.02)p 
trial 2 	1.78 x 10-4 = k (0.004)m (0.008)n (0.02)p  
trial 3 	1.78 x 10-4 = k (0.002)m (0.016)n (0.02)p 
trial 4 	3.56 x 10-4 = k (0.002)m (0.008)n (0.04)p 

r = k  [I‑] [BrO3-] [H+]

8.89 x 10-5 = k (0.002)(0.008)(0.02)
k  = 277.8 L2/mol2 h 

r = 277.8 L2/mol2 h  [I‑] [BrO3-] [H+]

r = 277.8 L2/mol2 h  [I‑] [BrO3-] [H+]
5.0 x10-4 = 277.8 L2/mol2 h  (0.0075)(0.0075) [H+]

[H+] = 0.0320 mol/L	trial 2 	1.78 x 10-4 = k (0.004)m (0.008)n (0.02)p  
trial 1 	8.89 x 10-5 = k (0.002)m (0.008)n (0.02)p 

(0.04/0.02)m = 2m = 2          m = 1

trial 3 	1.78 x 10-4 = k (0.002)m (0.016)n (0.02)p 
trial 1 	8.89 x 10-5 = k (0.002)m (0.008)n (0.02)p 

(0.016/0.08)n = 2n = 2          n = 1

trial 4 	3.56 x 10-4 = k (0.002)m (0.008)n (0.04)p 
trial 1 	8.89 x 10-5 = k (0.002)m (0.008)n (0.02)p 

(0.04/0.02)p = 2p =2             p = 1	

9.	     

10.	   

Zero order

11.	    
(a)	    
    [A]o = 0.841 mol/L
    [A]t = 0.25 mol/L
    t = 11.6 h
    k = 	t k = ln([A]o/[A]t)

(11.6 h) k = ln(0.841/0.25)

k = 0.105 h-1	
(b)	   
    t1/2  = ln 2/k = ln 2/0.105 h-1

    t1/2  = 6.63 h		
(c)	    
    [A]o = x
    [A]t = 0.61x
    t = 
    k =0.105 h-1 	t k = ln([A]o/[A]t)

t (0.105 h-1) = ln(x/0.61x)

t = 4.7 h	

12.	    
(a)  
    Ao = 10 g
    At = 
    t = 150 min
    k = 0.0277 min-1	t k = ln([A]o/[A]t)

(150 min)(0.0277 min-1) = ln(10/x)

x = 0.157 g 
	
(b)  
    Ao = 10 g
    At = 
    t = 180 min
    k = 0.0277 min-1	t k = ln([A]o/[A]t)

(180 min)(0.0277 min-1) = ln(10/x)

x = 0.068 g 
	
(c)   
       % = (At/Ao)x100 %
           = (0.068 g/10.0 g) 100 % 
           = 0.68 %
		
(d)    
    Ao = 9.20 mg
    At = 2.40 mg
    t = 
    k =0.0277 min-1 	t k = ln(Ao/At)

t (0.0277 min-1) = ln(9.2/2.4)

t = 48.5 min	

13.	  Both (a) and (b) 
Reason: 
•	Both have a molecularity < 2 in each step
•	Add to the overall rxn
•	Agree with the rate law