Workshop 4 - detailed answer key PDF

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General Chemistry 1

Workshop 4: The Chemical Equation and Bond Energy (10 pts) Part I. Balancing Chemical Equations Balancing chemical equations is a fundamental idea that you need to get very good at. Often when you are tasked with solving a problem in chemistry you need to start by writing and balancing the corresponding chemical equation. There are two skills required to write balanced chemical equations 1) being able to write down the correct chemical formulas for reactants and products, and 2) being able to balance the reaction by counting the atoms (or groups of atoms, like polyatomic ions) on either side of the equation. At the end of the day we want to balance chemical equations so that we can determine the relative ratios between reactants and products. From there you can determine, for example, how much product can be made in a chemical reaction. 1. To begin let’s practice writing and balancing some relatively simple chemical equations. Write and balance the following reactions. Designate the state of matter of each chemical by abbreviation in parenthesis: (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous solution – solution in water. Notice that if we say “solution” in general chemistry we always mean “aqueous solution.” a) Ammonia gas reacts with oxygen gas to yield nitrogen gas and liquid water. 4NH3(g) + 3O2(g)  2N2(g) + 6H2O(l) b) Tetraphosphorous decaoxide is a white crystalline solid reacts with water at room temperature to yield hydrogen phosphate solution (aka phosphoric acid). P4O10(s) + 6H2O(l)  4H3PO4(aq) c) At room temperature aluminum metal reacts with hydrogen bromide solution (aka hydrobromic acid) to yield aluminum bromide solution and hydrogen gas. 2Al(s) + 6HBr(aq)  2AlBr3(aq) + 3H2(g) d) Calcium chlorate solution is mixed with potassium phosphate solution to yield potassium chlorate solution and a white precipitate of calcium phosphate. 3Ca(ClO3)2(aq) + 2K3PO4(aq)  Ca3(PO4)2(s) + 6KClO3(aq) Notice that in this example you don’t need to balance phosphorus, chloride and oxygen atoms by themselves – phosphate ions and chlorate ions appear on both sides of the equation – so when balancing the equation for such reactions (they called double displacement reactions) – you should treat all polyatomic ions as single items.

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General Chemistry 1

2. Now let’s try a bit more complex problems. Consider the reaction of iron (II) sulfide with oxygen gas to form iron (III) oxide and sulfur dioxide. Write down and balance the equation. 4 FeS (s) + 7 O2(g)  2 Fe2O3(s) + 4 SO2 (g) ASSESS: Does your result make sense? Think about a way to show that the equation is balanced. E.g.: Reactants

Products

Fe = 4

Fe = 2 x 2 = 4

O = 2 x 7 = 14

O = (2 x 3) + (4 x 2) = 14

S=4

S=4

We are now ready to use our balanced chemical equation to answer the following questions. A. If you start with 100 formula units of FeS, how many SO2 molecules can you make? B. How many O2 molecules do you need to produce 2.89 x 1056 formula units of Fe2O3? PREDICT: In your group discuss the questions below and write down the answers. a) Do you think the number of SO2 molecules that will be produced in question A will be bigger or smaller than 100? _equals to 100____________ Please explain why: The ratio from equation is 4:4  1:1 ratio b) Do you think the number of O2 molecules that will be required in question B will be bigger or smaller than 2.89 x 1056? _bigger_________________ Please explain why: The ratio from equation is (7 O2)/(2 Fe2O3) = 3.5 SOLVE: Now answer the 2 questions above by setting up the appropriate unit conversions. A.

(100 FeS) *(4 SO2) = (100 SO2) (4 FeS)

B.

(2.89 x 1056 Fe2O3) * (7 O2) = (1.01 x 1057 SO2) (2 Fe2O3)

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General Chemistry 1

Part II. Bond Energies In this part you will bring together your knowledge of writing chemical formulas, writing and balancing chemical equations, and drawing Lewis structures to solve a bond energy problem. Solving this type of problems is not a trivial task and these questions often feel more difficult than they are – because of all the steps that you need to go through. We already practiced most of these steps – so all you need to do to solve these problems is to put together a plan and then carefully execute it. To practice, you will work in your group to solve the problem below. Start by planning the solutions and finish your work with the assessment of the result. Problem: Dichlorine monoxide is an unstable molecule and breaks readily in an exothermic process to form chlorine gas and oxygen gas. The heat produced in this reaction for each mole* of oxygen gas formed is ΔHrxn = -180.6 kJ/mol.* Use the table of bond energies from the next page to answer the following question: What is the bond energy (in kJ/mol) of the Cl---O bond in OCl2? The Lewis structure scaffolding for dichlorine monoxide is Cl-O-Cl. *Mole is a unit that measures the number of items in a specific sample. The symbol for the mole is “mol.” 1 mole of items is equal to 6.02 x 1023 items. You don’t need to worry about this unit for now – we will learn more about it in Topic 12. In that topic we will explain how chemists came up with this unit. At this point, you simply need to understand that mole is proportional to the number of items (molecules, atoms, ions, bond, etc.). Also notice that many energy values, like heats of reaction or bond energies, are measured for 1 mole of specific items. E.g. in our problem the heat of the reaction was measured in kJ/mol (per 1 mole of O2 gas formed) and you will report the Cl-O bond energy in kJ/mol (kilojoules per 1 mole of Cl-O bonds). Write your solution below. Please show and explain all steps. Plan:

1. Write and balance chem equation. 2. Draw Lewis structures for all molecules to determine types of bonds in all reactants and products. 3. Use ΔHrxn = ΔH bonds_broken + ΔHbonds_formed to find Cl---O bond energy

Solve: 1. Write down the chemical equation for the reaction: 2Cl2O  2Cl2 + O2

ΔHrxn = -180.6 kJ/mol

2.

So you have to break 4 O-Cl bonds and make 2 Cl-Cl bond and 1 O=O bond. 3. ΔHreaction = ΔH bonds broken + ΔH bonds formed - 180.6 = 4(x) + 2(-243) + 1 (-498) Notice that we need to use negative bond energy values for all formed bonds 4x = -180.6 + 486 + 498 x = 803.4/4 = 200.85 kJ/mol  200.9 kJ/mol or 201 kJ/mol 3

General Chemistry 1

Notice that bond energies in the table are all positive values. The bond energy (or bond enthalpy) – is the measure of bond strength and is defined as the average value of the bond dissociation energies for all bonds of the same type in the chemical species (usually at room temperature, 298K). In other words, to measure a bond energy we will need to measure how much energy will be spent to break this bond at 298K.

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