11/17 in-class activity

1.  What is the molar mass of HBr in SI units?

Molar mass of HBr = 80.908 g/mol, but THIS IS NOT IN SI UNITS!

In SI units, molar mass of HBR =  0.080908 kg/mol

2.  What are the SI units for the following quantities?

Pressure	pascal (Pa)
Volume	liter (L)
Mass	kg
Temperature	Kelvin (K)
Energy	Joule (J)

(J  and Pa are actually an SI-derived units, J is equal to kg*m²/s², and Pa is N/m²)

3. The mole fraction of argon in dry air is 0.00934.  How many liters of air at STP will contain enough argon to fill a 35.4-L cylinder to a pressure of 150. atm at 20^oC?

How many moles of Ar must there be to fill the cylinder?
PV=nRT

150. atm * 35.4 L = n * 0.08206 atm*L/mol*K * 293 K

n 220.8 mol Ar

STP: 0^oC = 273 K and 1 bar = 0.9869 atm

P_i = 0.00934 bar, P_tot = 1 bar (note we are using bar, not atm, because we are told we are at a total pressure of 1 bar).  P_i is the partial pressure of Ar in the atmosphere in bar.

We use the P_i to find the volume of air needed to obtain 220.8 mol:

P_iV=nRT
Note: WE DO NOT USE THE SAME R.  WE USE R IN UNITS OF BAR.  Conversely, we could convert P_i to atm and use the same R.  We must be sure that units cancel!

R = 0.08314 L*bar/mol*K

(Also, not the temperature change.)

0.00934 bar * V = 220.8 mol * 0.08314 L*bar/mol*K * 273 K

V = 536,665 L of dry air at STP

4. Which of these will change the position of equilibrium (circle all that apply)?

            allow more time to pass

            add a catalyst

            remove some products

Adding a catalyst does not shift the equilibrium (it only lowers the activation energy – more to follow on this later in the class).  Allowing more time to pass will not change a system at equilibrium from equilibrium.

5.  For the following reaction, predict the effect of each change to the system.

N₂ (g) + 6HCl (g) + energy ßà 2NH₃ (g) + 3 Cl₂ (g)

Triple the volume of the system:

Shifts left

The amount of nitrogen is doubled:

Shifts right

Heat is added to the system:

Shifts right

6.  Fill in the following table:

pH	pOH	[H+]	[OH-]	Acid, base, or neutral?
6.88	7.12	1.3 x 10-7	7.6 x 10-8	Acid
0.92	13.08	0.12	8.4 x 10-14	Acid
10.89	3.11	1.3 x 10-11	7.8 x 10-4	Base
7.00	7.00	1.0 x 10-7	1.0 x 10-7	Neutral

7.  What is the conjugate base of acetic acid?  Would you expect it to be a strong or weak base?  What is the conjugate acid of potassium hydroxide?  Would you expect it to be a strong or a weak acid?

Acetic acid (pK_a = 4.75): CH₃COOH ßà CH₃COO^-  (acetate anion)+ H⁺

Acetate anion is the conjugate base.  You would expect it to be a strong base.

KOH is a strong base, so you would expect its conjugate acid, K^-, to be a weak acid.

8.  Would you expect a sodium acetate solution to be acidic or basic?  Would you expect an ammonium chloride solution to be acidic or basic?

You would expect a sodium acetate solution to be basic (acetate is a strong base and sodium cation is a weak acid).

You would expect ammonium chloride solution to be acidic (ammonium is a strong acid and chloride anion is a weak base).

9.  The value for K_a is 7.45 x 10^-4 for citric acid (C₆H₁₀O₈) (a monoprotic acid).  Calculate the pOH for a 0.200 M citric acid solution.

We can write this equilibrium in terms of a “general acid” HA:

HA ßà A^- + H⁺

We also have the autoiniozation of water:

H₂O ßà H⁺ + OH^-

We know that the concentration of H⁺ in water is 1.0 x 10^-7 M, so we start out with 1.0 x 10^-7 M H⁺ concentration (I in the ICE diagram).  Generally, we won’t need to account for this, but sometimes we do.  We will include it just in case.

Concentrations (M)	HA	A-	H+
I	0.200	0	1.0 x 10-7
C	-x	+x	+x
E	0.200 - x	x	1.0 x 10-7 + x

We can write: K_a = [A^-][H⁺]/[HA]

Plugging in the equilibrium concentrations, we get:

7.45 x 10^-4 = x*(1.0 x 10^-7 + x)/(0.200 – x)

x² + 7.451 x 10^-4x – 1.49 x 10^-4 = 0

Plugging this into the quadratic equation, we get two values for x, but only one of them makes physical sense.  We get x = 0.01184 M.

Therefore, equilibrium concentration of H⁺ = 0.01184 M

pH = -log[H⁺] = 1.937

pOH = 14 – pH = 12.073

We can check our answer by plugging the equilibrium values into the equation for K_a.  If we get back the correct value, we know we have the correct answer.

Original worksheet:

1.  What is the molar mass of HBr in SI units?

2.  What are the SI units for the following quantities?

Pressure
Volume
Mass
Temperature
Energy

3. The mole fraction of argon in dry air is 0.00934.  How many liters of air at STP will contain enough argon to fill a 35.4-L cylinder to a pressure of 150. atm at 20^oC?

4. Which of these will change the position of equilibrium (circle all that apply)?

            allow more time to pass

            add a catalyst

            remove some products

5.  For the following reaction, predict the effect of each change to the system.

N₂ (g) + 6HCl (g) + energy ßà 2NH₃ (g) + 3 Cl₂ (g)

Triple the volume of the system:

The amount of nitrogen is doubled:

Heat is added to the system:

6.  Fill in the following table:

pH	pOH	[H+]	[OH-]	Acid, base, or neutral?
6.88
			8.4 x 10-14
	3.11
		1.0 x 10-7

7.  What is the conjugate base of acetic acid?  Would you expect it to be a strong or weak base?  What is the conjugate acid of potassium hydroxide?  Would you expect it to be a strong or a weak acid?

8.  Would you expect a sodium acetate solution to be acidic or basic?  Would you expect an ammonium chloride solution to be acidic or basic?

9.  The value for K_a is 7.45 x 10^-4 for citric acid (C₆H₁₀O₈) (a monoprotic acid).  Calculate the pOH for a 0.200 M citric acid solution.

11/10 in-class activity

For the reaction of solid carbon (C_(s)) and gaseous carbon dioxide (with the product gaseous carbon monoxide), K_p=1.50.

1.) Write the balanced chemical equation (remember this is an equilibrium).

                                          k₁

            C_(s) + CO_2,(g)    ßà       2 CO_(g)

                                     k_-1

2.) Write the forward and reverse rate laws in terms of partial pressure.  Come up with an expression for equilibrium.

Forward rate = k₁ (Pco₂)

Reverse rate = k_-1 (Pco)²

k₁ is the forward reaction rate constant and k_-1 is the reverse reaction rate constant.

Equilibrium constant, K_p = (Pco)²/(Pco₂)

Remember that solids and pure liquids are not included in ICE diagrams of equilibrium expressions (it doesn’t make sense to have a concentration or a partial pressure of a solid or pure liquid).

3.) For excess C_(s) and initial partial pressures Pco₂ = 0.56 atm and Pco = 0.14 atm, find Q_p, the reaction quotient.  Which way will the reaction proceed to equilibrium?

Q_p = (Pco)²/(Pco₂) = 0.035

Q<K so the reaction will proceed to the right (products) to reach equilibrium.

4.) Make an ICE diagram for this reaction using the partial pressures.

	CO2            à	2CO
I (initial)	0.56 atm	0.14 atm
C (change)	-x	+2x
E (equilibrium)	0.56 - x	0.14 + 2x

5.) Solve for x.  You will get two values for x here.  Which one makes physical sense?

K_p = 1.50 (given in the problem)

Given the equilibrium constant expression K_p = (Pco)²/(Pco₂), we can use the new equilibrium concentration expressions (found in the ICE table under equilibrium).  Plugging this in, we get:

1.50 = (0.14 + 2x)²/(0.56 – x)

multiplying both sides by (0.56 – x):

1.5*(0.56 – x) = (0.14 + 2x)²

0.84 – 1.50x = 0.0196 + 0.56x + 4x²

Therefore, 0 = 4x² + 2.06x – 0.8204

Plugging into the quadratic equation (see video), we get:

x = 0.263 atm and x = -0.778 atm

We know that the partial pressure of CO₂ will decrease, so x must be positive.  Also, for x = -0.778, we get an equilibrium pressure of CO of 0.14 + 2*-0.778, which doesn’t make physical sense (we can’t have a negative partial pressure).  Therefore, we get x = 0.263 atm.

6.) If you add 0.042 mol CO to the 2L system at 293 K at equilibrium, what is Q_p?  According to Le Châtelier’s principle, to which direction will the reaction proceed?

For x = 0.263 atm, we get equilibrium partial pressures of: Pco = 0.666 atm and Pco₂ = 0.297 atm.  Putting the added number of moles into the equation for the ideal gas law (PV=nRT), we get a partial pressure of 0.505 atm.  We add this to the partial pressure of CO we already have and plug it into the expression for Q_p:

Q_p = (0.505 + 0.666)²/(0.297) = 4.62

Q>K, so the reaction will proceed to the left (toward reactants).  Le Châtelier’s principle also tells us that when we add more products to a system at equilibrium, the system will shift toward reactants.

Remembering the quadratic equation

11/3 in-class activity

A cylinder of molecular hydrogen is separated by a metal plate from a cylinder of molecular oxygen.  The plate is removed, a spark starts the reaction, and gaseous water is formed.  Assume the product has had time to cool to ambient temperatures of 400K.  24 g of molecular hydrogen tracts with 76 g of molecular oxygen.  The volume of each cylinder is 1.00 L.  (Use the ideal gas law, not Van der Waal’s equation of state.)

1.) Write the balanced equation.

                                                   

2 H₂(g) + O₂ (g) à 2 H₂O (g)

2.) How many moles of molecular hydrogen and molecular oxygen are there originally?  What is the limiting reagent?

12 mol H₂, 2.375 mol O₂

O₂  is the limiting reagent.

3.) What is the initial pressure in each cylinder in atm (before mixing the reagents)?  Under these estimated pressures, would you expect the gases to behave ideally?

Use PV=nRT.  Use R = 0.08206 L atm/mol K so that the units cancel.

P for H₂ = 394 atm

P for O₂ = 77.96 atm

For pressures above 10 atm, you would not expect the gases to behave ideally.  These pressures are far higher than 10 atm, so you would not expect the gases to behave ideally.  However, the question indicates that you should use the ideal gas law, although it is likely a poor approximation in this situation.

4.) How many moles of water are formed?  How many moles of excess reagent are there?

4.75 mol H₂O formed

7.25 mol excess O₂

5.) What are the resulting partial pressures of water and the excess reagent in the combined cylinders? What is the total pressure?

Partial pressures: 78.0 atm H₂O, 119 atm H₂

Total pressure: 197 atm

*Note that each resulting number should have two significant figures.  Rounding should be done that the end here, but extra digits were included for comparison.

A fun way to remember the molecular formulas for polyatmics!