Reaction Rate and Rate Law

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Chemistry 521 Exam I, Spring Quarter
7:30 - 9:00 PM, 25 April 2000

NAME
1. [10 points] What are the concentrations of HSO− , SO2− , and H+ in a 0.20 M KHSO4 solution? (Hint: 4
4
H2 SO4 is a strong acid; Ka = 1.3 × 10−2 for HSO− .)
4

2. [15 points] Calculate the pH of 1.00 L of the buffer 1.00 M CH3 COONa/1.00 M CH3 COOH (pKa = 4.74) before and after the addition of (a) 0.080 moles NaOH and (b) 0.12 moles HCl. (Assume there is no change in volume).

3. [10 points] The following reaction is found to be first order in A: A −→ B + C
If half of the starting quantity of A is used up after 56 seconds, calculate the fraction that will be used up after 6.0 minutes.

4. [15 points] The rate law for the decomposition of ozone to molecular oxygen 3 O3 (g) −→ 3 O2 (g)
is
rate = k

[O3 ]2
.
[O2 ]

The mechanism for this process is
k1
O3

O + O2
k−1
k

2
O + O3 −→ 2 O2

Derive a rate law from these elementary steps. Clearly state the assumptions you use in the derivation. Explain why the rate decreases with increasing O2 concentration.

5. [15 points] Consider the following parallel reactions
k

1
A −→ B

k

2
A −→ C

The activation energies are 45.3 kJ/mole for k1 and 69.8 kJ/mole for k2 . If the rate constants are equal at 320 K, at what temperature will k1 /k2 = 2.00?

6. [15 points] The pre-exponential factor and activation energy for the unimolecular reaction CH3 NC(g) −→ CH3 CN(g)
are 4.0 × 1013 s−1 and 272 kJ/mole, respectively. Calculate the values of ∆S ◦‡ , ∆H ◦‡ , and ∆G◦‡ at 300 K.

7. [20 points] An enzyme-catalyzed reaction (KM = 2.7 × 10−3 M) is inhibited by a competitive inhibitor I (KI = 3.1 × 10−5 M). Suppose that the substrate concentration is 3.6 × 10−4 M. How much of an inhibitor is needed for 65% inhibition? How much does the substrate concentration have to be increased to reduce the inhibition to 25%?

Constants
R = 8.314 J-K−1 -mol−1
N0 = 6.022 × 1023 mol−1

kB = 1.381 × 10−23 J-K−1
h = 6.626 × 10−34 J-s

Equations
ax2

+ bx + c = 0

pH = pKa + log

and

x=

−b ±

me = 9.11 × 10−31 kg
−12 C2 N−1 m−2
0 = 8.854 × 10



b2 − 4ac
2a

[proton acceptor]
[proton donor]

Quadratic Equation
Henderson-Hasselbalch Equation

[A] = [A]0 − kt
ln

e = 1.1602 × 10−19 C
c = 2.998 × 108 m-s−1

Zeroth-order rate law

[A]
= −kt
[A]0

First-order rate law

t1/2 = ln 2/k

First-order Half Life

1
[A]

Second-order rate law

= kt +

1
[A]0

k = Ae−Ea /RT

Arrhenius Equation

1
1
1
+
=
µ
mA mB

Reduced mass

k=

kB T −∆G◦‡ /RT
he

Thermodynamic Formulation of reaction rate

k=

kB T ∆S ◦‡ /R −∆H ◦‡ /RT
e
he

Thermodynamic Formulation of reaction rate

Ea = ∆H ◦‡ − (∆n‡ − 1)RT
k = e kB T e∆S
h

◦‡ /R

k = e2 kB T e∆S
h
ν = νmax ·

Activation Energy and Free Enthalpy Relationship

e−Ea /RT

◦‡ /R

Unimolecular Thermodynamic Formulation of reaction rate

e−Ea /RT

Bimolecular Thermodynamic Formulation of reaction rate

[S ]
[S ] + KM

Michaelis-Menton Rate

[S ]

ν = νmax ·

[S ] + KM 1 +
ν = νmax 1 +

[I ]
KI

·

ν = νmax 1 +

[I ]
KI

·

Competitive Inhibition Rate

[I ]
KI

[S ]
[S ] + KM

Non-Competitive Inhibition Rate

[S ]
[S ] + KM 1 +

[I ]
KI

Uncompetitive Inhibition Rate

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