1 For the reaction 2 NO (g) + 2 H2(g) right arrow N2(g) + 2 H2O(g) At 1100 degre

Question

1 For the reaction 2 NO (g) + 2 H2(g) right arrow N2(g) + 2 H2O(g) At 1100 degree C, the following data have been obtained. Derive a rate law for the reaction and determine the value of the rate constant. 2 If the initial concentration of the reactant in a first order reaction A right arrow products is 0.64 mol/L and the half-life is 30.0 s, a) Calculate the concentration of the reactant exactly 60 s after initiation of the reaction. b) How long it would take for the concentration of the reactant to drop to one-eight its initial value? c) How long it would take for the concentration of the reactant to drop to 0.060 mol/L?

Explanation / Answer

1) rate of the reaction = k*[NO]x*[H2]y

Now, 0.012 = k*(5*10-3)x*(0.32)y...........(1)

0.024 = k*(1*10-2)x*(0.32)y.........(2)

(2)/(1) we get

2 = (2)x

or x = 1

Also, 0.096 = (1*10-2)x*(0.64)y,.........(3)

(3)/(2) we get

4 = (2)y

or, y = 2

Thus, k = 0.012/[(5*10-3)*(0.32)2] = 2.34375 M-2 L2 s-1

Hence the rate law becomes:- rate of reaction = k*[NO]*[H2]2 ; where k = rate constant

2) The governing equation of the first order reaction is :-

k*t = ln{[A0]/[A]} ; where k = rate constant and t = time in which the concentration reduces from A0 to A

thus, k = ln2/half life = 0.693/30 = 0.0231 s-1

(a) 0.0231*60 = ln{0.64/[A]}

or, [A] = 0.16 M L-1

(b) 0.0231*t = ln{1/(1/8)}

or, t = (1/0.0231)*ln8 = 90 sec

(c) 0.0231*t = ln(0.64/0.06)

or, t = 1024 sec

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