ODD (i.e.top) PROBLEMS:

1. The equilibrium constant for a reaction occurring in the gas phase is K_p = 1000 atm^-1. What will be the units of K?
   1. ^oC.
   2. nanometers.
   3. moles/liter.
   4. moles
   5. liter/moles
   ANSWER: The key here is to notice the units on K_p. Atm^-1 means that upon conversion to K, the units will be (moles/liter)^-1 or liter/moles, E.

2. For the reaction: H₂(g) + I₂(g) <===> 2HI(g) we have initial partial pressures of, P^o_I2 = 1.0 atm, P^o_HI = 1.0 atm, and P^o_H2 = 0 atm. What happens?
   1. Nothing.
   2. Reaction shifts left.
   3. Reaction shifts right
   4. Direction of shift depends on K_p.
   ANSWER: The reaction quotient, Q, is infinity for these intial conditions suggesting that no matter what the value of K_p, the reaction will proceed to the left as written. Now this shift might well be imperceptible because, for example, K_p could be very small. However in principle, the shift must be to the left, B.

3. Consider the fictional reaction: A(g) + 3B(g) <===> AB(g) + B2(g). Which statement is true?
   1. K_p = K
   2. K = K_p(RT)²
   3. K = K_p(RT)^-2
   4. K = K_p(RT)
   5. K = K_p(RT)^-1
   ANSWER: This conversion is discussed on p. 603 and 604 of Zumdahl. The change in the number of moles of gas, n, is -2 for this reaction (two moles of products - four moles of reactants = -2). Therefore, when we convert to K (using P/RT = n/V), we are left with a factor of RT², B.

4. Consider the fictional reaction: CD(g) + D(g) <===> C(g) + D₂(g). If all four species are present intially at partial pressures of 1.0 atm,, and K_p = 1 atm^-2, which statement is true?
   1. The reaction will shift to the right.
   2. The reaction will shift to the left.
   3. The reaction is at equilibrium.
   4. A & C.
   5. B & C.
   ANSWER: The reaction quotient, Q, equals unity and the value of K_p is unity. So these initial conditions describe equilbrium, C.

EVEN (i.e. bottom) PROBLEMS:

1. For the reaction: H₂(g) + I₂(g) <===> 2HI(g) we have initial partial pressures of, P^o_H2 = 1.0 atm, P^o_HI = 1.0 atm, and P^o_I2 = 0 atm. What happens?
   1. Nothing.
   2. Reaction shifts left.
   3. Reaction shifts right
   4. Direction of shift depends on K_p.
   ANSWER: The reaction quotient, Q, is infinity for these intial conditions suggesting that no matter what the value of K_p, the reaction will proceed to the left as written. Now this shift might well be imperceptible because, for example, K_p could be very small. However in principle, the shift must be to the left, B.

2. The equilibrium constant for a reaction occurring in the gas phase is K_p = 100 atm². What will be the units of K?
   1. ^oC.
   2. nanometers.
   3. (moles/liter)².
   4. (liters/mole)²
   5. moles²
   ANSWER: The key here is to notice the units on K_p. Atm² means that upon conversion to K, the units will be (moles/liter)², C.

3. Consider the fictional reaction: A(g) + 3B(g) <===> AB(g) + B₂(g). If all four species are present intially at partial pressures of 1.0 atm, and K_p = 1 atm^-2, which statement is true?
   1. The reaction will shift to the right.
   2. The reaction will shift to the left.
   3. The reaction is at equilibrium.
   4. A & C.
   5. B & C.
   ANSWER: The reaction quotient, Q, equals unity and the value of K_p is unity. So these initial conditions describe equilbrium, C.

4. Consider the fictional reaction: CD(g) + D(g) <===> C(g) + D₂(g). Which statement is true?
   1. K_p = K
   2. K = K_p(RT)²
   3. K = K_p(RT)^-2
   4. K = K_p(RT)
   5. K = K_p(RT)^-1
   ANSWER: This conversion is discussed on p. 603 and 604 of Zumdahl. The change in the number of moles of gas, n, is 0 for this reaction (two moles of products - two moles of reactants = 0). Therefore, K=K_p, A.