The rate of the reaction of nitrogen monoxide with oxygen was measured at 25 deg

Question

The rate of the reaction of nitrogen monoxide with oxygen was measured at 25 degrees C. starting with various initial concentrations of reactants NO and O2.

2NO(g) + O2(g)---> 2NO2(g)
initial concentrations reaction rate
mol/L mol/l.s
NO(g) O2(g)

Trial #1 .020 .010 .028
Trial #2 .020 .020 .057
Trial #3 .040 .020 .227


A. Calculate the rate constant for the reaction and reaction orders.

B.Write the complete rate equation using the calculated values for the rate constant and reaction order.

c. Calculate the reaction rate for this reaction at 25 degrees C, when the inital concentration of NO is 0.010 mol/l and that of O2 is 0.020 mol/l.

Please show the calculations on how you came to the following answers. I have alot allready figured out, but am missing some peices to complete my answers. Thank you

Explanation / Answer

let rate law be

r = k[NO]a[O2]b

(a)

for trial 1, we can write

0.028 = k[0.02]a[0.01]b

for trial 2 we can write

0.057 = k[0.02]a[0.02]b

take ratios of rates of trial 2 and 1

you will get

(57/28) = 2b

solving, b 1

similarly take ratios of rates of trial 2 and 3

(227/57) = 2a

solve, you will get a2

the rate law is

r = k[NO]2[O2]1

for specific rate constant

insert values of trial 1

0.028 = k[0.02]2[0.01]

so,

k = 7000 M-2s-1

hence the complete rate law is

r = 7000*[NO]2[O2]1

for final part

put [NO] = 0.01M

[O2] = 0.02 M

solving you will get rate = 0.014 mol/l.s

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