Homework Problem Set 1 Solutions

Transcription

1 Chemistry Dr. Jean M. Standard omework Problem Set 1 Solutions 1. A student investigates a bond between atoms A and B in a molecule using a software package for molecular mechanics. The student measures the bond length as 1.5 Å. When the single point energy of the molecule is calculated, the student gets a value of 10.0 kj/mol. The equilibrium A-B bond length is 1.00 Å. a. Using the information obtained above, determine the force constant for bond stretching. Use units of kj mol 1 Å. We start from the equation for the stretching energy, U s = 1 k s ( r r eq ). The equation may be solved for the force constant k s, Substituting, k s = k s = U s ( r r eq ). ( 10.0 kj/mol) ( Å), k s = 3840 kj mol 1 Å. 1 b. Plot the stretching energy as a function of bond length for this system (use Microsoft Excel or some other plotting or spreadsheet software package, or else sketch the expected behavior). Using the stretching force constant as k s = 3840 kj mol 1 Å, the stretching energy can be plotted for the bond.

2 a. What happens to the plot from question 1 if the stretching force constant is doubled? From the plot, if the force constant is doubled, the curvature of the stretching function increases (that is, the plot becomes narrower). b. What happens to the plot from question 1 if the stretching force constant is halved? If the force constant is halved, the curvature of the stretching function decreases (that is, the plot becomes broader).

3 3 3. In most molecular mechanics software packages, the bond lengths and other geometrical parameters of a molecule are determined using the cartesian coordinates of each atom. For example, the bond distance r AB between two atoms A and B can be calculated from the equation r AB = [( x A x B ) + ( y A y B ) + ( z A z B ) where (x A, y A, z A ) are the cartesian coordinates of atom A and (x B, y B, z B ) are the cartesian coordinates of atom B. Consider a water molecule with atoms having the following cartesian coordinates (in Å): Atom x y z O For this molecule, calculate the bond distances between pairs of atoms O- 1 and O-. The O bond lengths must be determined from the cartesian coordinates. Using the formula given, O bond length 1 is r O1 = [( x O x 1 ) + ( y O y 1 ) + ( z O z 1 ) [( ) + ( ) + ( ) r O1 = r O1 = 0.885Å. The length of O bond is [( ) + ( y O y ) + ( z O z ) [( ) + ( ) + ( ) r O = x O x r O = r O = 0.885Å.

4 4. The water molecule in problem 3 is assumed to obey a simple harmonic stretching and bending force field with the following parameters: 4 k O = 800 kj mol 1 Å r O,eq = 0.93 Å k O = 400 kj mol 1 radian θ O,eq = a. What is the stretching energy of water at this geometry? (Include both bonds in the calculation.) Substituting the lengths of the O bonds into the expression for the stretching energy, U s = 1 k O ( r O1 r O1,eq ) + 1 k O r O r O,eq = kj mol 1 Å ( ) ( )( Å) + 1 ( 800 kj mol 1 Å )( Å) U s = 5.67 kj/mol. 4 b. What is the bending energy for water at this geometry? For the bond angle, the bending energy is defined as U b = 1 k O ( θ O θ O,eq ). The bending angle must be determined from the cartesian coordinates. For example, the law of cosines may be employed, r 1 = r O1 + r O r O1 r O cosθ O. Using the figure below, the distance r 1 may be evaluated from cartesian coordinates. 1 θ O O

5 4 b. continued 5 The 1 - distance is [( ) + ( y 1 y ) + ( z 1 z ) [( ) + ( ) + ( ) r 1 = x 1 x r 1 = r 1 = Å. Substituting the 1 - distance into the law of cosines and solving yields The bending energy is therefore cosθ O = r O 1 cosθ O = cosθ O = , + r O r 1, r O1 r O ( 0.885Å) Å ( ) ( Å) ( )( 0.885Å) 0.885Å θ O = 1.65 radians or 94.6 o. E b = 1 ( 400 kj mol 1 rad ) 1.65rad 1.8 rad E b = 5.78 kj/mol. ( ),, 4 c. What is the total energy for water at this geometry? The total energy of the water molecule is the sum of the stretching and bending energies: U tot = U s + U b, U tot = 5.67 kj/mol kj/mol, U tot = 11.45kJ/mol.

6 5 a. Sketch a plot of the force field for rotation of ethane about its C-C axis. The position in which the hydrogens are eclipsed corresponds to a torsional angle of 0 (Fig. 1), while the position in which the hydrogens are staggered corresponds to 60 (Fig. ). Your sketch should show the energy of ethane as the torsional angle varies from 0 to Figure 1. Eclipsed conformation of ethane Figure. Staggered conformation of ethane The energy will be large due to steric interactions between hydrogen atoms at angles of 0, 10, 40, and 360 degrees. The energies of the eclipsed geometries, since they are all structurally identical, will be the same. The energy will be lowest due to alleviation of steric interactions between the hydrogen atoms at angles of 60, 180, and 300 degrees. The energies of these staggered geometries will all be the same since they are identical structurally. This will lead to a sinusoidal energy curve as shown below. (Note that we are not given what the energy difference is between the peaks and valleys for this part of the problem; hence, the y-axis scale is in arbitrary units.)

7 7 5. b. What is a simple functional form that would have the same qualitative shape as your plot in part a? The curve shown in part (a) looks like a cosine function. owever, since ethane has three-fold symmetry, it must be cos(3ω), where ω is the torsional angle. So, a reasonable force field would be: U t = V 3 [ 1 + cos(3ω) ], where V 3 is a constant that scales the height of the peaks (using the same notation as in the notes). c. If the maxima in the torsional potential of ethane occur at 1. kj/mol and the minima occur at 0.0 kj/mol, construct a specific potential function to reproduce the ethane torsional potential. The term 1 + cos(3ω) ranges from 0 to because the factor 1 shifts the entire cosine function up. In order to set the range of the potential to be 1. kj/mol from minima to maxima (rather than ), the cosine function must be scaled by a factor of 0.6. This sets the constant V 3 = 1. kj/mol. The overall potential function for ethane rotation would therefore be given by: U t = 0.6 [ 1 + cos(3ω) ].

8 6. Explain how your plot from question 5a would change if the molecule was instead 1,-dichloroethane. 8 For dichloroethane, the plot of energy as a function of rotation angle is very similar to the one sketched in problem 5 for ethane. owever, it is a little different because at 0 and 360 degrees, the two chlorine atoms are eclipsed, while at 10 and 40 degrees, the chlorine atoms are eclipsed by hydrogen atoms. So, the energy at 0 and 360 degrees is higher than the energy at 10 and 40 degrees. Similarly for the staggered forms, the two chlorine atoms are furthest apart at 180 degrees, so the energy is lowest for that configuration than for 60 or 300 degrees. The plot from problem 5 must therefore be modified to look something like the plot shown below. Notice that this looks qualitatively like the plot for n-butane where instead of methyl groups, 1,-dichloroethane has chlorine atoms. Energy Angle (degrees)