Solutions to Homework Assignment #5
CHM 152
Spring 2002
13.59
Higher temperatures may disrupt the intricate three dimensional structure of the enzyme, thereby reducing
or totally destroying its catalytic activity.
13.64
First, calculate the radius of the 10.0 cm sphere.
3
V =
4 3
πr
3
4 3
πr
3
3
10.0 cm =
r = 1.34 cm
The surface area of the sphere is:
2
2
2
area = 4πr = 4π(1.34 cm) = 22.6 cm
3
Next, calculate the radius of the 1.25 cm sphere.
V =
4 3
πr
3
4 3
πr
3
3
1.25 cm =
r = 0.668 cm
The surface area of one sphere is:
2
2
2
area = 4πr = 4π(0.668 cm) = 5.61 cm
2
2
The total area of 8 spheres = 5.61 cm × 8 = 44.9 cm
2
Obviously, the surface area of the eight spheres (44.9 cm ) is greater than that of one larger sphere (22.6
2
cm ). A greater surface area promotes the catalyzed reaction more effectively.
13.81
(a)
(b)
(c)
(d)
(e)
14.3
The equilibrium constant expression is the mathematical expression of the law of mass action. The law of
mass action states that for a reversible reaction at equilibrium and a constant temperature, a certain ratio of
reactant and product concentrations has a constant value, K, (the equilibrium constant).
14.9
a)
b)
A catalyst works by changing the reaction mechanism, thus lowering the activation energy.
A catalyst changes the reaction mechanism.
A catalyst does not change the enthalpy of reaction.
A catalyst increases the forward rate of reaction.
A catalyst increases the reverse rate of reaction.
Kc =
Kc =
[NH3 ]2
[NO2 ]2 [H 2 ]7
[SO2 ]2
[O2 ]3
Kp =
Kp =
P2
NH3
2
PNO
⋅ PH7
2
2
PSO
PO3
2
2
2
14.10
2
PCO
PCO2
c)
Kc =
[CO]2
[CO 2 ]
d)
Kc =
[C6 H5 COO − ][H + ]
[C6 H5 COOH]
Kp =
K p = K c ( RT ) ∆n
Kp is the equilibrium constant expressed in terms of gas pressure.
Kc is the equilibrium constant expressed in terms of molar concentration.
R is the ideal gas constant (0.0821 L⋅atm/mol⋅K)
T is the temperature in Kelvin
∆n = moles of gaseous products − moles of gaseous reactants
14.13
When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant
becomes the reciprocal of the original equilibrium constant.
K' =
14.14
1
1
=
= 2.40 × 1033
−34
K
. × 10
417
The problem states that the system is at equilibrium, so we simply substitute the equilibrium concentrations
into the equilibrium constant expression to calculate Kc.
Step 1: Calculate the concentrations of the components in units of mol/L. The molarities can be
calculated by simply dividing the number of moles by the volume of the flask.
2.50 mol
= 0.208 M
12.0 L
[H 2 ] =
[S2 ] =
135
. × 10 −5 mol
= 113
. × 10 −6 M
12.0 L
[H 2S] =
8.70 mol
= 0.725 M
12.0 L
Step 2: Once the molarities are known, Kc can be found by substituting the molarities into the equilibrium
constant expression.
Kc =
[ H 2S]2
2
[ H 2 ] [S 2 ]
=
(0.725) 2
−6
2
(0.208) (113
. × 10 )
= 1.08 × 10 7
If you forget to convert moles to moles/liter, will you get a different answer? Under what circumstances
will the two answers be the same?
14.15
Using Equation (14.5) of the text: Kp = Kc(0.0821 T)
∆n
where, ∆n = 2 − 3 = −1
and T = (1273 + 273) K = 1546 K
22
Kp = (2.24 × 10 )(0.0821 × 1546)
−1
20
= 1.76 × 10
14.17
We can write the equilibrium constant expression from the balanced equation and substitute in the
pressures.
Kp =
2
( 0.050)2
PNO
=
= 0.051
. )( 0.33)
PN 2 PO2 ( 015
Do we need to know the temperature?
14.18
The equilibrium constant expressions are:
(a)
Kc =
(b)
Kc =
[NH3 ]2
[N 2 ][H 2 ]3
[NH3 ]
1/ 2
[N 2 ]
[H 2 ]3 / 2
Substituting the given equilibrium concentration gives:
(a)
Kc =
(b)
Kc =
(0.25) 2
(011
. )(191
. )3
= 0.082
(0.25)
(011
. )
1/ 2
(191
. ) 3/2
= 0.29
Is there a relationship between the Kc values from parts (a) and (b)?
14.20
Because pure solids do not enter into an equilibrium constant expression, we can calculate Kp directly from
the pressure that is due solely to CO2 (g).
Kp = PCO2 = 0.105
Now, we can convert Kp to Kc using the following equation.
Kp = Kc(0.0821 T)
Kc =
Kp
(0.0821T )∆n
Kc =
14.21
∆n
.
0105
(0.0821 × 623)
(1− 0)
= 2.05 × 10 −3
We substitute the given pressures into the reaction quotient expression.
Qp =
PPCl3 PCl2
PPCl5
=
( 0.223)( 0111
. )
.
= 0140
( 0177
. )
The calculated value of Qp is less than Kp for this system. The system will change in a way to increase Qp
until it is equal to Kp. To achieve this, the pressures of PCl3 and Cl2 must increase, and the pressure of
PCl5 must decrease.
Could you actually determine the final pressure of each gas?