Quiz 2, with solutions

Useful equations provided with the quiz included $PV = nRT$ and the quadratic formula.

The phosphorus transformation

At $T = 1200\,^\circ\text{C}$ the reaction

$$\text{P}_4(g) \rightleftharpoons 2\,\text{P}_2(g)$$

has an equilibrium constant $K = 0.6$.

(a) Suppose the initial partial pressure of $\text{P}_4$ is 2.00 atm and that of $\text{P}_2$ is 2.00 atm. Calculate the reaction quotient $Q$ and state whether the reaction proceeds to the right or to the left as equilibrium is approached.

Solution. The reaction quotient is given by

$$Q = \frac{P_{\text{P}_2}^2}{P_{\text{P}_4}}$$

Plugging in the given numbers, we get $Q = 2$. We have $Q > K$, and therefore the equilibrium shifts to the left.

(b) Calculate the partial pressure of $\text{P}_2$ at equilibrium.

Solution. We know that the total initial pressure is 4 atm. After equilibrium, the total pressure remains at 4 (because the number of particles remains the same). Let the partial pressure of $\text{P}_2$ at equilibrium be $x$. We have $P^{eq}_{\text{P}_2} = x$ and $P^{eq}_{\text{P}_4} = 4 - x$. Plugging into the equilibrium constant, we solve

$$K = 0.6 = \frac{x^2}{4-x} \implies \mathbf{x = 1.278}$$

(c) If the volume of the system is then increased, will there be net formation or net dissociation of $\text{P}_4$?

Solution. When the volume increases, the total pressure decreases. To “counter” this change, the system shifts to the side with more particles, giving net dissociation of $\text{P}_4$.

Fun with morphine

Morphine is a weak base for which $K_b$ is $8\times 10^{-7}$. Denote morphine by M.

(a) Write the equilibrium expression of morphine in water.

Solution. A weak base behaves in water as

$$\text{M} + \text{H}_2\text{O} \rightarrow \text{HM}^+ + \text{OH}^-$$

(b) Calculate the pH of a solution made by dissolving 0.0400 mol of morphine in water and diluting to 600.0 mL.

Solution. The initial concentration of morphine is $[\text{M}] = 0.04/0.6 = 0.067$. Setting up an “ICE” table, we obtain

$$K_b = \frac{[\text{OH}^-][\text{HM}^+]}{[\text{M}]} = \frac{x^2}{0.067 - x}$$

Solving for $x$ (or approximating the denominator as 0.067, just make sure you know the conditions when you are allowed to do so), we get $\mathbf{x = 0.00023}$.

Bonus, midterm revisited

Only attempt one of the following questions.

1. Examine the following Hamiltonian and determine what species of atom it describes.

$$\widehat{H} = \frac{-\hbar^2}{2m_e}\big(\nabla_1^2 + \nabla_2^2\big) - \frac{3e^2}{4\pi\epsilon_0}\Big(\frac{1}{r_1} + \frac{1}{r_2}\Big) + \frac{e^2}{4\pi\epsilon_0 r_{12}}$$

Choices are H, He, He⁺, Li, Li⁺.

Solution. The second term describes the nuclear-electron attraction potential. Comparing it to Coulomb’s law, we see that the atomic number $Z$ is 3, and there are 2 electrons. Therefore the answer is Li⁺.

2. Which of the following orbital occupations will allow for free rotation (e.g. cis-trans isomerization) of a bond?

1. $\sigma_{p_z}^2\, \pi_{p_x}^2\, \pi_{p_y}^0\, \pi_{p_x}^{*0}\, \pi_{p_y}^{*0}\, \sigma_{p_z}^{*0}$
2. $\sigma_{p_z}^2\, \pi_{p_x}^1\, \pi_{p_y}^1\, \pi_{p_x}^{*0}\, \pi_{p_y}^{*0}\, \sigma_{p_z}^{*0}$
3. $\sigma_{p_z}^2\, \pi_{p_x}^1\, \pi_{p_y}^0\, \pi_{p_x}^{*1}\, \pi_{p_y}^{*0}\, \sigma_{p_z}^{*0}$
4. $\sigma_{p_z}^0\, \pi_{p_x}^0\, \pi_{p_y}^0\, \pi_{p_x}^{*2}\, \pi_{p_y}^{*2}\, \sigma_{p_z}^{*0}$

Solution. For free rotation to occur in a conjugated system, the double bond needs to be broken. We look for the option where the $\pi$ character is gone while the sigma character remains, which is option 3.