1/2 N2O4 (g) ------> NO2(g) delta H = 28.6 kJ at 30 degrees C, k = 5.1 x 10^6/s,

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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