The reaction between nitrogen dioxide and carbon monoxide is
NO2(g)+CO(g)→NO(g)+CO2(g)
The rate constant at 701 K is measured as 2.57 M−1⋅s−1 and that at 895 K is measured as 567 M−1⋅s−1. The activation energy is 1.5×102 kJ/mol.
Predict the rate constant at 525 K .
Express the rate constant in liters per mole-second to three significant figures.
___________________________ L/mol*s
Solution :-
Given data
K1 = 2.57 M-1s-1
T1 = 701 K
K2= ?
T2 = 525 K
Ea = 150 kJ/ mol * 1000 J */ 1 kJ = 150000 J / mol
Formula to calculate the rate constant using the activation energy is as follows
ln[K2/K1] = (Ea/ R )*[(1/T1)-(1/T2)]
where R= 8.314 J per mol. K
lets put the values in the formula and calculate the activation energy
ln [K2/2.57] = 150000 J per mol / 8.314 J per K . mol * [(1/701)-(1/525)]
ln [K2/2.57] = -8.628
K2 / 2.57 = anti ln [-8.628]
K2/2.57 = 0.000179
K2 = 0.000179 *2.57
K2 = 0.00046
Therefore rate constant at 525 K is 4.6*10^-4 L/mol. s
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