# 4 R = 8.31451 J mol–1 K–1 = 0.0820578 L atm mol–1 K–1 = 1.9872 ```Chemistry 231 Third Exam, Version 1
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R = 8.31451 J mol –1 K–1 = 0.0820578 L atm mol–1 K–1 = 1.9872 cal mol–1 K–1 = 0.0831451 L bar mol–1 K–1
N 0 = 6.022137 &times; 10 23 mol–1
g = 9.80665 m s–2
1.00 atm = 760 torr = 1.01325 &times; 105 Pa
F = 96 485.3 C/mol
c = 2.99792458 &times; 108 m/s
h = 6.62608 &times; 10–34 J s
e = 1.602177 &times; 10–19 C
D. (20) Transport. Thermal conductivity, viscosity, diffusion, and electrical conductivity are the four
transport properties we studied. For each of these, (1) give the fundamental transport equation we used;
(2) tell what quantity is transported (&quot;flow&quot;) and what gradient drives the transport; and (3) give the SI units
for the &quot;flow,&quot; the gradient, and the transport coefficient.
Chemistry 231 Third Exam, Version 1
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E. (20) Mechanisms and derivations. DO EITHER 1 OR 2, NOT BOTH.
1.
As discussed in class, the simplest mechanism for enzyme catalysis is a two-step one. In the first step
the enzyme (E) and substrate (S) combine to form a complex (ES); in the second, the complex proceeds
to product (P), releasing the enzyme catalyst.
(a) Give the equations of this mechanism in chemical form.
(b) Letting k 1 and k –1 represent the forward and reverse rates in the first step, and k 2 and k –2 the same
for the second step, apply the steady-state approximation to ES to obtain an expression for [ES] in
terms of the 4 k's, [S], [P], and [E].
(c) Unfortunately, [E] is generally not known during the course of the reaction, though [E]0 is.
Replace [E] by its equivalent in terms of [E]0 and [ES] in your initial steady-state equation, and re-solve
for [ES]SS .
(d) Use your result from (c) to obtain an expression for the initial rate r0 ≡ r at t = 0, where r =
d[P]/dt. Rewrite your result in terms of the Michaelis constant KM (clearly defined). {Hint: Set [P] =
0 and [S] = [S] 0.}
Chemistry 231 Third Exam, Version 1
2.
assumptions.
(b) Stokes' Law for the frictional force on a solid sphere of radius r moving at speed υ through a
Newtonian fluid of viscosity η is F fr = 6πηrυ. At terminal speed in a gravitational field, this force is
equal and opposite to the gravitational force. Use this relationship to solve for the terminal speed in
terms of the quantities already defined, the gravitational acceleration g, and the densities of the sphere
and fluid, ρ and ρ fl, respectively.
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Chemistry 231 Third Exam, Version 1
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F . (10) Electrolytes. Consider an aqueous solution of K3PO4 at a concentration of 0.055 m˚.
1.
2.
Calculate ν &plusmn; for this solute.
Write a balanced chemical equation for the formation of the ion pairs in this solution.
3.
Suppose that the molality of free phosphate anion in solution is 0.019 m˚. Calculate the molalities of all
other solute species, and determine α.
Calculate the ionic strength of this solution.
4.
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