Question

The reaction between nitrogen dioxide and carbon monoxide isNO2(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.

Answer 1

Answer:

Explanation:

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