26 Marks: 1 For the combustion of ethanol with oxygen, suppose molecular oxygen

Question

26 Marks: 1 For the combustion of ethanol with oxygen, suppose molecular oxygen is reacting at the rate of 1.668 M/s. At what rate (in M/s) is ethanol reacting? Answer: 27 For the combustion of ethanol with oxygen, suppose molecular oxygen is reacting at the rate of 2.263 M/s. At what rate (in M/s) is water being formed? Marks: 1 Answer 28 For the reaction of nitric oxide (NO) and oxygen to form nitrogen dioxide, if molecular oxygen is reacting at the rate of 0.00066 M/s, what is the rate (in M/s) of nitric oxide reacting? Marks: 1 Answer: 29 Marks: 1 For the reaction of nitric oxide (NO) and oxygen to form nitrogen dioxide, if molecular oxygen is reacting at the rate of 0.00728 M/s, what is the rate (in M/s) of nitrogen dioxide being formed? Answer. 30 Marks: 1 If the initial rate is 0.0558 Mis for a reaction with a rate law of ratekA, what is the value of the rate constant when the initial concentration of [A] is 0.0781 M? Answer:

Explanation / Answer

26.

The reaction equation is expressed as

C2H5OH(l) + O2(g) ----> CO2(g) + H2O(g)

The balanced reaction equation is

C2H5OH(l) + 3 O2(g) ----> 2 CO2(g) + 3 H2O(g)

Rate = -d[C2H5OH] / dt = -1/3 [d[O2]/dt] = ½ d[CO2]/dt = 1/3 d[H2O]/dt

But,

= - [d[O2]/dt] = 1.668 M/s

So,

-d[C2H5OH] / dt = -1/3 [d[O2]/dt]
-d[C2H5OH] / dt = 1/3 (1.668 M/s)
                              = 0.556 M/s

27.

The reaction equation is expressed as

C2H5OH(l) + O2(g) ----> CO2(g) + H2O(g)

The balanced reaction equation is

C2H5OH(l) + 3 O2(g) ----> 2 CO2(g) + 3 H2O(g)

Rate = -d[C2H5OH] / dt = -1/3 [d[O2]/dt] = ½ d[CO2]/dt = 1/3 d[H2O]/dt

But,

= - [d[O2]/dt] = 2.263 M/s

-1/3 [d[O2]/dt] = 1/3 d[H2O]/dt

[d[O2]/dt] = d[H2O]/dt

d[H2O]/dt = 2.263 M/s

28.

The reaction equation is expressed as

NO + O2 ----> NO2

The balanced reaction equation is

2NO + O2 ----> 2NO2

Rate = -1/2 d[NO] / dt = - d[O2]/dt] = 1/2 d[NO2]/dt

But,

= - [d[O2]/dt] = 0.00066 M/s

Since, - d[O2]/dt] = -1/2 d[NO]/dt

0.00066 M/s = -1/2 d[NO]/dt

- d[NO2]/dt = 2 x 0.00066 M/s = 0.00132 M/s

29.

The reaction equation is expressed as

NO + O2 ----> NO2

The balanced reaction equation is

2NO + O2 ----> 2NO2

Rate = -1/2 d[NO] / dt = - d[O2]/dt] = 1/2 d[NO2]/dt

But,

= - [d[O2]/dt] = 0.00728 M/s

Since, - d[O2]/dt] = 1/2 d[NO2]/dt

0.00728 M/s = 1/2 d[NO2]/dt

d[NO2]/dt = 2 x 0.00728 M/s = 0.01456 M/s

30.

rate = k[A]

or k = rate / [A]

Initial rate = 0.0558 M/s
Initial concentration of [A] = 0.0781 M

So, rate constant, k = (0.0558 M/s ) / (0.0781 M)
= 0.7145 s-1

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