1.SOLUTIONS-ANSWERS

a) õ = k[A] í = 1 first order reaction

b) õ = k[A][B] í =1+1=2 second order reaction

c) õ = k[A][B]

*2**.**Solution*

Due to the fact that the reaction is simple, the
rate is given by

õ
= k[A][B]^{2}^{ }
* *
(1)

a) If
we double the concentration of A the rate becomes

õ_{1} = k[2A][B]^{2 }=
2k[A][B]^{2}
(2)

From
(1) and (2) we have õ_{1 }=
2õ.
Therefore the rate of the reaction is doubled.

b) If we double the concentration of B the
rate becomes

õ_{2} = k[A][2B]^{2 }=
4k[A][B]^{2}
(3)

From
(1) and (3) we have õ_{1} = 4õ.
Therefore the rate of the reaction is quadrupled.

c) If
we dilute the solution until it becomes twice its original volume, the
concentrations decrease to one half their initial value. The rate low gives us:

õ_{3} = k[0.5A][0.5B]^{2 }=
0.125k[A][B]^{2}
(4)

From
(1) and (4) we get õ_{3} =
0.125õ = õ/8.
Therefore the rate of the reaction decreases to one
eighth its initial value.

*3. *
*Solution*

From the reaction we can see that 0.001 moles of
HCl are consumed.

*4. Answer*

a)
0.025 mol/L·s, 0.025 mol/L·s b)zero.

* * *
Return to dictation*

*5. Answer*
* *á) 0.025 mol/L·s and
0.011 mol/L·s

â)

*Return
to dictation*

*6**.*
* Answer*

Either
2 or 8.

*
Return to dictation*

*7* *Solution*

The
analogy 1:2 means that, if the quantity of A is a moles, B would be 2a moles.

The evolution of the reaction is described as
follows:

From the
stoichiometry of the reaction and the initial quantities,results that the
reactants are fully consumed. Thus,

á - x = 0 =>á = x. From this
result we conclude that ,after the termination of the reaction the quantities of
A and B are fully consumed. So, the curves are diminishing and have a tendancy
to the zero axis 0 (t).

Therefore, the demanded diagram is:

*Return
to dictation*

*8. Answer*

á)

â)
0.0125 mol/L^{.}s

* *
*Return
to dictation*

* 9. Answer*

a 2, b 6, c 3, d 5.

*Return
to dictation*

* 10. Answer*

á) 0.02 mol/L·s, 0.01 mol/L·s, 0.005
mol/L·s, 0.0025 mol/L·s, 0.009375 mol/L·s,

â)

*Return
to dictation*

*11*. *Solution*

The progress of the reaction is described as
follows:

a) 0.2 mol, 0.1 mol and 0.1 mol

â)

*Return
to dictationç*

*12. *
*Solution
*The rate is given from the type:

So, the demanded rates of consumption and production
are:

õ = -Ä[Á] / Ät => -Ä[Á] = õ·Ät = 1,6·10^{-3}·50
= 0,08 mol/L
So, the total quantity of A that is consumed is
2,5^{.}0,08 = 0,2 mol.

Therefore, the composition of the mixture after the
elapse of the 50 seconds is:

0,4 mol Á, 0.3 mol
Â and 0.4 Ã.

*
Return to
dictation*

*13.
Solution
*
Due to the fact that we don’t
know whether the reaction is simple (one step only) or not and we don’t have
any other relevant information, we ignore the way it was inscribed earlier and
we write the rate low as follows:

õ
= k[A]^{x}[B]^{y
}
From this type we get the following equations for
each one of the measurements.

4·10^{-3}
= k 0.1^{x}0.1^{y
}(1)

8·10^{-3}
= k 0.2^{x}0.1^{y}
(2)

1.6·10^{-2}
= k 0.1^{x}0.1^{y
}(3)

From (1) and (2) we get:

From (1) and (3) we get:

a)
The rate low is given from the type õ = k[A][B]^{
2}

b) The reaction’s order is í = 1+2 = 3 (Third order)

c) From (1) we get:

*14.*
Answer

b)
í = 3, c) 0.5L^{3}
mol^{-}^{3}·s^{-}^{1},
d) 7.5·10^{-}^{3}
mol·L^{-}^{1}·sec

* * *Return
to dictation*

b) í=4,
c) k=9L^{3}·mol^{-3}·s^{-1
}^{
}*Return
to dictation*
^{
}

We
know that the relative pressure of a gas, component of a mixture, is given by
the equation P=[A]RT, which gives us [A]=Pa/RT. From this relationship we can
see that the relative pressure of a gas constitutes a way to measure the gas’s
concentration in the gas mixture. According to the above, the arithmetical
values of the relative pressures of the reactants, are equal to the relative
values of their concentrations. Therefore the rate low, for the above reaction,
it can also be written in the following form: õ
= k·P_{A}^{x
}P_{B}^{y}
where k is a constant with the suitable units (dimensions).

As we did in the solution of exercise 3, in order to write down the
rate low we ignore the reaction’s coefficients, because we don’t know from
the dictation whether the reaction is simple (one step only) or not. According
to the above expression of the rate low, we get for each one of the
measurements:

1.25·10^{-}^{3}
= k 0.2^{x}0.25^{y
}(1)

2.5·10^{-}^{3}
= k 0.4^{x}0.125^{y
}(2)

5·10^{-}^{3}
= k 0.4^{x }0.25^{y
}(3)

from (1) and (3) we get:

from (2) and (3) we get:

The
rate low is given by: õ = k·P_{A}^{2}P_{B}

The reaction is third order because í = 2+1= 3

í = 2, k = 5·10^{-3}
mol·L^{-1
}s^{-1 }atm^{-}^{2}
^{
}*Return
to dictation*

*18* *Answer*

a) í = 0, b) mol/L·sec

*Return
to dictation*

*19*. *Solution*

The initial concentrations are:

The progress of the reaction appears in the
following table

2ÍÏ +
O 2ÍÏ_{2}

0.3
0.2 0
(initial mol/L)

0.2
0.1
0.2 (mol/L
that react and produced)

0.1
0.1 0.2
(final mol/L)

From dictation we know that the reaction is simple
(only one step), which dives us:

õ
= k[NO]^{2}[O_{2}].

This type for the initial situation becomes:

For the final situation we have:

õ_{2}
= 0.2·0.1^{2 }0.1
= 2·10^{-}^{4}
mole·L^{-}^{1}s^{-}^{1}

*20*. *Answer*

0.45 , 0.45 and
0.9 mol

* 21. Solution*

á) 6.4
/ 80 = 0.08 mol of Á.

The progress
of the reaction is described as follows:

From equation (1) for P=1.64 atm we
get 0.02 moles. So, at the specific time we have 0.04 moles of A, 0.04 moles of
B and 0.02 moles of C. The rate is given by the type:

õ_{1}
= k(0.04/3)^{2}
(2)

Also, from equation (1) for P=1.804 atm we get 0.03 moles. Therefore at
the specific time we have 0.02 moles of A, 0.06 moles of B and 0.03 moles of C.
the rate is given by the type:

õ_{2}
= k(0.02/3)^{2}
(3)

From (2) and (3) we get: õ_{1}:õ_{2}
= 4:1

â)

* * *Return
to dictation*

^{ }