31/ 2NO + H2 ----> N2O + H2O experiment initial[NO] (mol/L) initial [H2] ( mol/L

Question

31/ 2NO + H2 ----> N2O + H2O

experiment initial[NO] (mol/L) initial [H2] ( mol/L) rate ( mol/L*S)

1 6.4 x 10^-3 2.2 x 10^-3 2.6 x 10^-5

2 12.8 x 10^-3 2.2 x 10^-3 1.0 x 10^-4

3 6.4 x 10^-3 4.5 x 10^-3 5.1 x 10^-5

I/ write the rate law for this reaction

II/ Determine the rate constant for this reaction with the appropriate units/

III/ what is the order of this reaction?

47/ of the following, a 0.10M aqueous solution of __________ will have the highest boiling point.

A/ NACL

B/ AL(NO3)3

C/ K2CrO4

D/ Na2SO4

48/ Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction:

2NO2 -----> 2NO + O2

In a particular experiment at 300 celcius , [NO2] drops form 0.0100 M to 0.00650 M in 100 seconds. The rate of appearance of O2 for this period is _________ M/s.

A/ 1.8 x 10^-5

B/ 3.5 x 10^-5

C/ 7 x 10^-5

D/ 3.5 x 10^-3

E/ 7 x 10^-3

Explanation / Answer

31) Given data

2NO + H2 ----> N2O + H2O

experiment

Run    initial[NO] (mol/L)             initial [H2] ( mol/L)           rate ( mol/L*S)

1           6.4 x 10^-3                            2.2 x 10^-3                2.6 x 10^-5

2          12.8 x 10^-3                           2.2 x 10^-3               1.0 x 10^-4

3           6.4 x 10^-3                           4.5 x 10^-3                 5.1 x 10^-5

----------

Let the rate of reaction be represented as,

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

with k being the rate constant

First take run 1 and run 2

The concentration of H2 is constant here and gets cancelled,

2.6 x 10^-5/1.0 x 10^-4 = k[6.4 x 10^-3]^x/k[12.8 x 10^-3]^x

0.26 = 0.5^x

Taking log on both sides,

log(0.26) = xlog(0.5)

x = 2.0

So the reaction is second order with respect to NO

Now take run 1 and 3, here concentration of NO remaing constant and gets cancelled out,

2.6 x 10^-5/5.1 x 10^-5 = k[2.2 x 10^-3]^y/k[4.5 x 10^-3]^y

0.51 = 0.50^y

Taking log on both sides,

log(0.51) = ylog(0.50)

y = 1

So the reaction is first order with respect to H2 concentration

The rate law becomes,

I. rate = k[NO]^2[H2]

II. rate constant k = 2.6 x 10^-5/(6.4 x 10^-3)^2 (2.2 x 10^-3) = 288.53

III. the order of reaction = 2+1 = 3

47) of the following, a 0.10M aqueous solution of B) Al(NO3)3 will have the highest boiling point.

Since the concentration remaing the same for all, the boiling point would depend on the number of ions present in solution.

In A) two ions Na+ and Cl-, B) 4 ions, C) 3 ions and in D) 3 ions are present in solution. Thus, B) will have the highest boiling point of all.

48) Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction:

2NO2 -----> 2NO + O2

In a particular experiment at 300 celcius , [NO2] drops form 0.0100 M to 0.00650 M in 100 seconds. The rate of appearance of O2 for this period is A) 1.8 x 10^-5 M/s.

From the reaction we can see that, 2 moles of NO gives 1 mole of O2.

amount of NO reacted = 0.01 M - 0.0065 M = 0.0035 M

So the amount of O2 formed would be = 0.0035/2 = 0.00175 M

rate of O2 formation = 0.00175 / 100 = 1.8 x 10^-5 M/s

A) 1.8 x 10^-5

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