Rates of Reaction (and thermodynamics)

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27 Οκτ 2013 (πριν από 3 χρόνια και 7 μήνες)

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Chemical Equilibrium

Chapter 17

Review

Given a chemical equation…


A + B


C + D


Which letters represent the products?


Which letters represent the reactants?

What is a rate?

What is a concentration?

Reversible Reactions


A


B


B


A



Reversible reactions occur in both the
forward and reverse directions

CH
4

+ 2H
2
S


CS
2

+ 4H
2

Means that CH
4

+ H
2
S can form CS
2

+ H
2
or CS
2

+ H
2
can form CH
4

+ H
2
S

Or A

B

Equilibrium


When


A


B




B


A


The reactions are at
equilibrium
when the
forward and reverse reactions balance
each other because they take place at
equal rates




If A

B

r1

r2

r1 = r2

equilibrium

Equilibrium


Amounts of products and reactants stays
constant


Does not mean that there are equal amounts
of products and reactants: [A] does not have
to equal [B]


Catalysts speed both forward and backward
rates, so they do not affect equilibrium
concentrations


Law of chemical equilibrium: at a given
temperature, reactant and product
concentrations have a constant value

Equilibrium Constants

For a reversible reaction at equilibrium

aA + bB


cC + dD

K
eq
=

[C]
c

x [D]
d




[A]
a

x [B]
b


Exponents same as

coefficients in equation

Brackets indicate

concentration

Products over

reactants

If <1, more reactants
than products

If >1, more products
than reactants

Writing expressions for
Homogeneous Equilibria


All in the same state of matter


Follow formula on previous slide

Ex. Write the equilibrium constant expression for
CH
4
(g) + 2H
2
S(g)


CS
2
(g) + 4H
2
(g)


K
eq

=


[CH
4
][H
2
S]
2

[CS
2
][H
2
]
4

Writing expressions for
Heterogeneous equilibria


Different states of matter


Because solids and liquids have a
concentration that doesn’t change, you
can omit them from the equilibrium
expression

Ex. What is the equilibrium constant expression
for MgCO
3
(s)


MgO(s) + CO
2
(g) ?


[MgCO
3
]

[MgO][CO
2
]

Keq=

= [CO
2
]

Calculating equilibrium constants


Plug in concentration data (experimental)
into K
eq

expression

Ex. At a certain temperature, the following
concentrations are measured. Calculate Keq
if
CH
4
(g) + 2H
2
S(g)


CS
2
(g) + 4H
2
(g) and


[CH
4
]=0.5 mol/L

[CS
2
]=0.5 mol/L


[H
2
S]=0.7 mol/L


[H
2
]=0.8 mol/L



[CH
4
][H
2
S]
2

[CS
2
][H
2
]
4

Keq=

(0.5)(0.7)
2

(0.5)(0.8)
4

=0.84

Calculating K
eq


Calculate Keq for the following reaction:



2SO
2

+ O
2



2SO
3

If there are 0.1 mol , 0.2 mol, and 0.3 mol
in a 1.0 L flask.

More on equilibrium constants


K
eq

should always equal the same thing for a
given reaction at a given temperature


An infinite number of equilibrium positions:
varies depending on initial concentration


Must be a closed system: no reactant or
product can be added or escape


Temperature must remain constant


Dynamic: forward and reverse reactions don’t
stop

Le Chatelier’s Principle


If a stress is applied to a system in
dynamic equilibrium, the system
changes to relieve the stress


3 main stresses


Changing concentration of reactants or
products


Changing the temperature


Changing the volume or pressure


Le Chatelier’s Principle


Concentration


2SO
2

+ O
2



2SO
3



Increasing the concentration of a reactant
will shift the equilibrium so that more of it
is used up

Decreasing the concentration of a product
will shift the equilibrium so that more of it
is made

Shifts reaction this way

Le Chatelier’s Principle


Pressure

2SO
2

(g)+ O
2
(g)


2SO
3
(g)



If pressure is increased (or volume decreased), it
favors the side with fewer moles of gas

If pressure is decreased (or volume increased), it
favors the side with more moles of gas

Shifts reaction this way

If we increase the pressure…

If we increase the volume…

Le Chatelier’s Principle


Temperature

2SO
2

+ O
2



2SO
3
+ heat




If heat is added, reaction will shift to favor the
endothermic reaction
(you can treat heat like a
chemical) . If the forward reaction is exothermic
and T increases, Keq decreases.


Shifts reaction this way

If we increase the temperature…

If we decrease the temperature…

Using equilibrium constants


If you know an equation and Keq, you
can calculate equilibrium concentrations.

Ex.
If 2ICl

I
2

+ Cl
2

K
eq
=0.110 at a particular temperature,
[I
2
]=0.0330, and [Cl
2
]=0.220, what is the
[ICl]?

[
ICl
]
2

[
I
2
][
Cl
2
]

K
eq

=


[
ICl
]
2

(0.033)(0.22)

0.110

=


[
ICl
]=0.257 mol/L

Solubility Product Constants


Recall that compounds vary in solubility

At saturation…

AgCl(s)


Ag
+
(aq) + Cl
-
(aq)

So we can write a K
eq

for this process:

K
eq
=[Ag+ ][Cl
-
]



[AgCl]


K
sp
=[Ag
+

][Cl
-
]




(multiply both sides by [AgCl])


=1.7 x 10
-
5


Solubility product constant

Solubility Products

K
sp
=[Ag
+

][Cl
-
]

= 1.7 x 10
-
5

What is the concentration of each ion at
saturation?

Since

AgCl(s)


Ag
+
(aq) + Cl
-
(aq)


[Ag
+
]=[Cl
-
]=x

So,

x
2
= 1.7 x 10
-
5


Predicting Precipitates and
Calculating Ion Concentrations


To determine saturation you can do a
trial solubility product (Qsp) and compare
it to Ksp

Ex. Will CaF
2

precipitate if equal volumes
0.020 M Ca(NO
3
)
2

and 0.0064 M NaF
are mixed at 298 K? Ksp for CaF
2
=3.5 x
10
-
11

Qsp=[Ca
2+
][F
-
]
2
=(0.01)(0.0032)
2
=1.0 x10
-
7



Qsp > Ksp


Yes

Note: concentrations are reduced by half because
the volume of solution is now 2x as big

The Common Ion Effect


The solubility of a compound will
decrease if you mix it with a solution with
the same ion.


A precipitate will form


Ex: When mixing BaSO
4

and NaSO
4
,
BaSO
4

will precipitate out

The Common Ion effect


What is the concentration of [Br
-
] if we
add 0.020 mol AgNO3 to 1.00 L of
saturated AgBr?