The Following data was collected for the first order reaction: 2N 2 O 5 = 2N 2 O
ID: 879580 • Letter: T
Question
The Following data was collected for the first order reaction:
2N2O5 = 2N2O4 + O2
Slope of the Regression Line = -11805.16 Y intercept= 52.37 Correlation factor= 0.9856 Ea= -98.15 kJ/mol
1a) Using this information, calculate the rate constant at 35 degrees C
1b) Using this data, at what temperature (degrees C) will the rate constant be 8.00 x 106 sec-1?
Temp (Celcius) Rate Constant k(sec-1) ln k Temp (K) 1/Temp (K-1) 0.0 7.87 x 103 8.97 273 .0037 25.0 3.46 x 105 12.75 298 .0033 45.0 4.98 x 106 15.42 318 .00314 65.0 4.87 x 107 17.70 338 .00296Explanation / Answer
Log(K2/K1) = Ea/2.303R[1/T1-1/T2]
T1 = 25 C , T2 = 35 C
K1 = 3.46 x 10^5 k2 = ? Ea = 98.15 Kj/mol
Log(K2/3.46 x 10^5 )= 98.15 /2.303*8.314[1/298-1/308]
K2 at 35 C = 1.252*10^6 sec-1
1b.
T1 = 25 C , T2 = ?
K1 = 3.46 x 10^5 k2 = 8.00 x 10^6 Ea = 98.15 Kj/mol
Log(8.00 x 10^6/3.46 x 10^5 )= 98.15 /2.303*8.314[1/298-1/T2]
T2 = 324.136 K
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