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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