Consider the the gas-phase decomposition ofNO2: NO2 --> NO + 1/2 O2 Temperature

Question

Consider the the gas-phase decomposition ofNO2:           NO2 --> NO + 1/2 O2

Temperature              Initial [NO2](M)                  Initial rate (M·s–1)
    600K                      0.0010                                 5.4 × 10–7
    600 K      0.0020                                 2.2 × 10–6
    700K                      0.0020                                 5.2 × 10–5

From this data, determine:

1)The activation energy Ea
2) If 0.0050 moles of NO2 are sealed in a 1.0-liter flask andallowed to decompose at 650 K, how long
would it take before the concentration of [NO2] = 0.0010 M?

Explanation / Answer

a) Formula :                 ln (k1/k2) = Ea /R ( 1/T2 -1/T1 ) From the table ,                          k1 = 2.2 × 10 ^–6Ms^–1
                         k2 = 5.2 × 10^–5Ms^–1                          T1 = 600 K                          T2 = 700 K     Upon substituting in the above formula, ln (2.2 × 10 ^–6 / 5.2 × 10^–5) = Ea / 8.314 J / mol.K ( 1/700 K - 1/ 600 K)                                 -3.16           = Ea ( -100 ) / 8.314 J / mol. * 700 * 600 K                                               Ea = 3.16 *  8.314 J / mol. * 700 * 600K / 100                                                     =  110.34 kJ / mol    
b ) It is a second order reaction,      Formula :                        1/[A] = kt + 1/ [A]0 [A]0 = 0.005 M [A]   = 0.0010 M k   at 650 K can be obtained by taking theequation   ln (k1/k2) = Ea /R ( 1/T2 -1/T1 ). Now upon substituting in the above formula time can beobtained.
    

   [A]0 = 0.005 M [A]   = 0.0010 M k   at 650 K can be obtained by taking theequation   ln (k1/k2) = Ea /R ( 1/T2 -1/T1 ). Now upon substituting in the above formula time can beobtained.

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