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1.For the reaction: 2NO(g) + 2H 2 (g) N 2 (g) + 2H 2 O(g) the following data at

ID: 894894 • Letter: 1

Question

1.For the reaction:     2NO(g) + 2H2(g)      N2(g) + 2H2O(g)       the following data at 1280 ºC was obtained:

            Exp #               [NO]                             [H2]                              initial rate (M/s)

            1                    5.0 x 10-3 M                  2.0 x 10-3 M                  1.3 x 10-5 M

            2                    1.0 x 10-2 M                  2.0 x 10-3 M                  5.0 x 10-5 M

            3                    1.0 x 10-2 M                  4.0 x 10-3 M                  1.0 x 10-4 M

Determine the rate law, the rate constant, and the reaction rate expected when   [NO] = 4.8 x 10-3 M and

[H2] = 6.2 x 10-3 M.

2. A certain first order reaction has an activation energy of 83 kJ/mol and a rate constant of 2.1 x 10-2 s-1 at 150. ºC. What is this reaction’s rate constant at 300.ºC

answers:

1. rate = k[NO]2[H2] , k = 260 M-2s-1 , rate = 3.7 x 10-5 M/s

2. 10 s-1

Explanation / Answer

1. Let us find order of the reaction,

let rate of reaction is represented as,

rate = k[NO]^x[H2]^y

with,

k = rate constant

x and y are order with respect to NO and H2

consider Exp# 1 and 2 : concentration of H2 are same for both so,

rate1/rate2 = 1.3 x 10^-5/5.0 x 10^-5 = (5 x 10^-3/1 x 10^-2)^x

0.26 = 0.5^x

taking log on both sides,

log(0.26) = xlog(0.5)

x = 2

So order is 2 with respect to NO

Now take Exp# 2 and 3 : Here concentration of NO is constant so,

rate2/rate3 = 5 x 10^-5/1 x 10^-4 = (2 x 10^-3/4 x 10^-3)^y

0.5 = 0.5^y

taking log on both sides and solving for y,

y = 1

So order is 1 with respect to H2

the rate law thus becomes,

rate = k[NO]^2[H2]

Now let us calculate rate constant from say Exp# 1 values,

rate = 1.3 x 10^-5 = k(5 x 10^-3)^2(2 x 10^-3)

k = 260 M-2.s-1

now when, [NO] = 4.8 x 10^-3 M and [H2] = 6.2 x 10^-3 M

rate = 260(4.8 x 10^-3)^2(6.2 x 10^-3) = 3.7 x 10^-5 M/s

2. Usng Arrhenius equation,

ln(k2/k1) = Ea/R[1/T1-1/T2]

we have,

k1 = 2.1 x 10^-2 s-1 at 150 oC

k2 = unknown at 300 oC

T1 = 150 oC = 150 + 273 = 423 K

T2 = 300 oC = 300 + 273 = 573 K

Ea = 83 kJ/mol = 83000 J/mol

R = 8.314 J/K.mol

Feed values,

ln(k2/2.1 x 10^-2) = 83000/8.314[1/423 - 1/573]

k2 = 10.1 s-1 (or 10 s-1)

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