The decomposition of nitrogen dioxide, 2 NO_2 (g) right arrow 2 NO (g) + O_2 (g)

Question

The decomposition of nitrogen dioxide, 2 NO_2 (g) right arrow 2 NO (g) + O_2 (g) has a rate constant of 0.498 M s^-1 at 3 19 degree C and a rate constant of 1.81 M s^-1 at 354 degree C. What are the values of the activation energy and the frequency factor for this reaction? What is the rate constant at 420 degree C? (Gas constant = 8.31 J K^-1 mol^-1) If the molar solubility of beryllium(II) hydroxide is 8.6 Times 10^-7 M in pure water, what is its K_sp value? What is the molar solubility of beryllium(II) hydroxide in a solution that is 1.50 M in NH_3 and 0.25 M in NH_4CI? (K_b of NH_3 = 1.8 x 10^-5)

Explanation / Answer

Rate constant, k1 = 0.498 Ms-1

Temperature, T1 = 319 DegC = 319+273 = 592 K

Now applying Arrhenius equation

k1 = A x (exp)-Ea/RT1

=>0.498 Ms-1 = Ax(exp)-Ea/8.31JK-1mol-1x592K -------- (1)

Here Ea = activation energy

A = Frequency factor

Rate constant, k2 = 1.81 Ms-1

Temperature, T1 = 354 DegC = 354+273 = 627 K

Now applying Arrhenius equation

k2 = A x (exp)-Ea/RT2

=> 1.81 Ms-1 =  A x (exp)-Ea/8.31 JK-1mol-1x627K ------- (2)

Dividing equation(1) and equation (2) we get

0.498 / 1.81 = Ax(exp)-Ea/8.31JK-1mol-1x592K / A x (exp)-Ea/8.31 JK-1mol-1x627K   

=> 0.498 / 1.81 = (exp)-Ea/8.31JK-1mol-1x592K / (exp)-Ea/8.31 JK-1mol-1x627K

=> ln(0.498 / 1.81) = - Ea/8.31JK-1mol-1x592K - (- Ea/8.31JK-1mol-1x627K)

=> - 1.29 =  - Ea/8.31JK-1mol-1x592K + Ea/8.31JK-1mol-1x627K

=> 1.29 = Ea / 8.31JK-1mol-1 x [ 35 K / 592 K x 627K]

=> Ea = 113687 J/mol = 113.7 KJ/mol (answer)

Now putting the value of Ea in equation (1) we get

0.498 Ms-1 = Ax(exp)-113687Jmol-1 / 8.31JK-1mol-1x592K  

=> A = 5.413 x 109 Ms-1 (answer)

At 420 DegC = 420+273 = 693 K, the rate constant can be calculated as

k = A x (exp)-Ea/RT

=> k =   5.413 x 109 Ms-1 x  (exp)-113687Jmol-1/8.31JK-1mol-1x693K

=> k = 14.5 Ms-1 (answer)

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