1/2 N2O4 (g) ------> NO2(g) delta H = 28.6 kJ at 30 degrees C, k = 5.1 x 10^6/s,
ID: 818356 • Letter: 1
Question
1/2 N2O4 (g) ------> NO2(g) delta H = 28.6 kJ
at 30 degrees C, k = 5.1 x 10^6/s, and the activation energy id 54.0 kJ/mol
USE THIS INFO FOR THE 3 questions
1.) this reaction is
A) one-half order
B) first order
C) second order
D) exothermic
2.) the rate constant at 45 degrees C is
A) 5.1 x 10^6/s
B) 1.4 x 10^7/s
C) 5.2 x 10^7/s
D) 8.3 x 10^8/s
E) 1.1 x 10^38/s
3.) what is the activation energy for the reverse reaction?
A) 25.4 kJ/mol
B) 28.6 kJ/mol
C) 32.8 kJ/mol
D) 54.0 kJ/mol
E) 82.6 kJ/mol
PLEASE SHOW STEPS! I will only award points to someone who shows thier work.
Explanation / Answer
1.
B. First order (Since rate constant k = 5.1 x 10^6/s. uint is s-1)
2. We know that ,
ln K2/K1 =Ea/ R (1/T1 -1/T2)
Where,
K2 = ? rate constant at temperatureT2 = 45 + 273 = 318 K
and K1 = 5.1 x 10^6/s rate constant at temperature at T1 = 30+ 273 = 303 K
Ea = activation energy = 54.0 kJ/mol = 54 x 10^3 J/mol
and R = 8.314 J/K/mol
by putting all these values in the equation....
K2/K1= e ^[54 x 10^3/ 8.314 (1/303 - 1/318)]
K2/K1 = e^[1.0392]
= 2.83 x K1
=> K2 = 2.83 x 5.1 x 10^6
= 1.44 x 10^7 s-1
B) 1.4 x 10^7/s
3. It will be same for both forward and backward reaction.
D) 54.0 kJ/mol
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