Download Problem set 1

CHEM-UA 127: Advanced General Chemistry I
Problem set #1: due 9/21
Practice problems from Chapter 1: 5, 11, 15, 39
Practice problems from Chapter 2: 31, 35, 36,52
Graded problems
1. A given element X has an average relative mass A and three isotopes of relative masses A1 , A2 and A3 . Let the
fractional abundances of these isotopes be p1 , p2 and p3 . Suppose the mass difference between isotope 2 and
isotope 1 is 1.003 (relative mass), with isotope 2 being the heavier than isotope 1, and that the mass difference
between isotope 3 and isotope 2 is 1.005 (relative mass), with isotope 3 being heavier than isotope 2. In addition,
suppose that isotope 1 is 10 times as abundant as isotope 2 and that isotope 2 and isotope 3 together comprise
15% of the total amount of X. Calculate the value of A in terms of the mass of the most abundant isotope of X.
2. Balance the following reactions. You may use any method you like.
a. Ca5 (PO4 )3 F + H2 SO4 + H2 O −→ Ca(H2 PO4 )2 ·H2 O + CaSO4 ·2H2 O + HF
b. Ca5 (PO4 )3 F + SiO2 + C −→ CO + CaO· 32 SiO2 + SiF4 + P4
+
c. BrO−
3 + CH2 (COOH)2 + H −→ BrCH(COOH)2 + CO2 + H2 O
Hint: The charge must be balanced as well, which means that the net charge (total positive charge − total
negative charge) must be the same on both sides of the reaction.
d. NH4 ClO4 + Al −→ Al2 O3 + N2 + HCl + H2 O
−
+
e. CO2 + NH+
4 + O2 −→ C5 H7 O2 N + NO2 + H2 O + H
Hint: Charge must be balanced as well, meaning that the net total charge must be the same on both sides.
Is there more than one solution?
3. Iron (Fe) can be produced from iron oxide (Fe2 O3 ) by reaction with carbon-monoxide gas (CO) according to:
Fe2 O3 (s) + 3CO(g) −→ 2Fe(s) + 3CO2 (g)
When 7.56 kg of CO gas are combined with 15.02 kg iron oxide in a furnace, it is observed that 9.54 kg of iron
are produced. What is the percentage yield of this reaction?
4. Let r denote the distance between the two atoms in a diatomic molecule AB. A realistic expression for the
potential energy V (r) stored in the molecule as a function of r is
2
V (r) = D0 1 − e−b(r−r0 )
where D0 , b and r0 are constants.
a. Sketch the plot of V (r) as a function of r. Remember that r > 0.
1
b. Derive an expression for the magnitude of the force F (r) between the two atoms as a function of r and
sketch the plot of F (r) vs. r.
c. If the distance between the atoms A and B is initially r0 , determine an expression in terms of D0 , r0 and
b for the work needed to dissociate the molecule, i.e. to move the two atoms to infinite separation.
d. If r is close to r0 , then the potential energy V (r) can be approximated by the quadratic form
V (r) ≈
1
2
k (r − r0 )
2
where k = V ′′ (r0 ). so that the force F (r) becomes a simple Hooke’s Law force F (r) ≈ −k (r − r0 ). By
computing V ′′ (r0 ), determine an explicit expression for k in terms D0 , r0 , and b, and using your expression
for k, plot the quadratic approximation for V (r) on the same graph you generated in part a for the
true/exact V (r).
5.
a. Masses of atoms and molecules are often determined by the technique of mass spectrometry, which is based
on the nature of the motion of charged particles in a magnetic field.
A charged particle moving with velocity v in a constant, uniform magnetic field H will experience a force
F due to the field that is perpendicular to the directions of both v and H (see figure below).
H
y
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
v
r.
x
x
x
x
x
F
z
x
x
x
x
x
x
x
x
x
x
x
x
x
x
FIG. 1.
This force causes v to change in time, so that both v and F are changing. In the figure, the magnetic field
is directed into the plane of the page. The particle is initially at a position r, moving initially to the right
with velocity v so that the force is initially down.
Simple geometrical vector addition can be used to deduce the motion of the particle. Recall that two
vectors v1 and v2 can be added geometrically by placing the tail of one vector at the head of the other.
The vector that completes the triangle is the sum v1 + v2 as shown below:
2
v1
v1
v2
+ v2
FIG. 2.
Given an initial velocity v(0) and force F(0), the velocity v(∆t) a short time ∆t later can be calculated
approximately by assuming the force is constant over the time interval. This means that the velocity at
time ∆t will be given in terms of v(0) and F(0) by the formula:
v(∆t) = v(0) +
∆t
F(0)
m
and the new position r(∆t) is given approximately by
r(∆t) = r(0) + ∆tv(0)
Given the new velocity v(∆t), the new force F(∆t) can be determined, and these quantities fed in as new
“initial conditions” to the above procedure. By iterating in this fashion, positions and velocities can be
computed approximately at times ∆t, 2∆t, 3∆t,...
Using simple geometrical vector addition and the above iterative procedure, deduce the nature of the of the
motion of the charged particle in the magnetic field H. Show all of your work and give a brief description
of the particle’s trajectory.
b. A particle of charge q moving in a region of uniform magnetic field H experiences a force whose magnitude
is qvH. This force will give rise to a centripetal acceleration so that, by Newton’s second law:
qvH =
mv 2
R
where R is the radius of the particle’s orbit, m is its mass, and v is the magnitude of its velocity upon
entering the region of the magnetic field.
Consider a charged particle, initial at rest, that is accelerated through an electric field E = (E, 0, 0) over
a distance l. The particle then enters a region of uniform magnetic field H perpendicular to its velocity.
Show that its mass m is given by
m=
where R is the radius of its orbit.
3
qH 2 R2
2El