PO1-You should be able to define and give examples of the following terms and concepts:
PO2-You should know that in an exothermic reaction the value of DH is negative, and that in an endothermic reaction the value of DH is positive.
Example 1- The equation for the combustion of methane is . When this reaction occurs, heat is produced along with carbon dioxide and water. Specifically, 213 kcal of heat are produced for every mole of methane consumed; DH = -213 kcal/mol. This reaction is exothermic. The products contain less energy than the reactants.
Example 2- The reaction requires an input of 31.3 kcal for every mole of carbon consumed; DH = +31.3 kcal/mol. This reaction is endothermic. The products contain more energy than the reactants.
Self-test 1- Select the diagram that depicts an exothermic reaction.
Self-test 2- Select the number that corresponds to the enthalpy change, DH, of the reaction represented by this diagram:1 2 3
PO3- You should recognize that you may use DH as a conversion factor, and you should be able to use it as such to calculate the amount of heat released when a given quantity of reactant is converted to product.
Example 1- The value of DH for the combustion of methane is -213 kcal/mol. If you were to burn 2.6 mol of methane, the reaction would release 553.8 kcal;
Example 2- A calorimeter is a device that chemists use to measure the amount of heat released in a chemical reaction. In one of her physical chemistry experiments, Polly Ester used a calorimeter to determine that burning an unknown amount of methane released 125 kcal of heat. She used this information to calculate that she had burned 0.587 mol of methane:. The minus sign associated with these heats indicates that energy was released.
Example 3- Here's an example of a calculation that involves two conversion factors, DH and MW. The MW of methane is 16.0 g/mol. How much heat would be released if you were to burn 25.0 g of methane? First you have to convert the mass of methane to its equivalent quantity in moles. Then you follow the same approach illustrated in Example 1 to calculate the amount of heat that would be released by burning this amount of methane:
Self-test 3- How many kcal of heat are released when 0.154 mol of methane burns? Enter your answer to 3 significant figures. kcal
Self-test 4- To 3 significant figures, how many moles of methane would you have to burn to generate 2500 kcal of heat? mol
Self-test 5- To 3 significant figures, if you burned 8.5 g of methane, how many kcal of heat would be produced? kcal
Self-test 6- To 3 significant figures, how many grams of methane would you have to burn in order to generate 500 kcal of heat? g
PO4- You should be able to write an expression for Keq for any reaction.
Example 1- The expression for the reactionis .
Example 2- The expression for Keq for the reaction is .
Self-test 7- Select the correct expression for Keq for the reaction .
Self-test 8- Select the reaction for which .
PO5- You should be able to do simple calculations to determine the amounts of reactants and products present at equilibrium when you are given the value of Keq for a reaction.
Example 1- There are two isomers of 2-butene, Z-2-butene and E-2-butene. Keq for the interconversion of these isomers at 25oC is 3.2:.
If you start with a sample containing 1000 molecules of pure Z-2-butene and heat it at 25oC until the system comes to equilibrium, how many molecules of E-2-butene will there be in the mixture? The stoichiometry of the balanced equation tells you that for every molecule of Z-2-butene that reacts, one molecule of E-2-butene is formed. Let n be the number of molecules of Z-2-butene that react. Then at equilibrium, 1000-n molecules of Z-2-butene will be left and n molecules of E-2-butene will have been formed. Keq = n/1000-n = 3.2. Multiplying both sides of this equation by (1000-n) gives n = 3.2(1000-n), which expands to n = 3200 - 3.2n. Combining terms containing n (i.e. adding 3.2n to each side of the equation) yields 4.2n = 3200. Therefore, n = 761.90.... Since the numbeer of molecules must be an integer, we round this up to 762, which means that there are (1000-762) = 238 molecules of Z-2-butene left at equilibrium. To check that this is correct, calculate the value of Keq using these values of n and 1000-n: Keq = 762/238 = 3.2016.... = 3.2. Note, too, that 762 + 238 = 1000, the number of molecules we started with.
There is another way to think about this problem. When we say that the value of Keq = 3.2, it means that the ratio of products to reactants is 3.2/1. The sum of the numerator and the denominator of this ratio is (3.2 + 1) = 4.2. Therefore, at equilibrium the E-2-butene will comprise (3.2/4.2)*100 = 76.2% of the mixture. In other words, ot the 1000 molecules in the mixture, (3.2/4.2)*1000 = 721.904..... or 722 will be E-2-butene and the remainder will be the Z-isomer.
Self-test 9- At 227oC Keq = 1.65 for the isomerization of Z-2-butene and E-2-butene is. If you start with 2000 molecules of Z-2-butene, how many molecules of E-2-butene will there be when the system comes to equilibrium?
Self-test 10- If Keq = 1.65 for the isomerization of Z-2-butene and E-2-butene, what percentage of the mixture will be E-2-butene when the system comes to equilibrium?. Enter your answer to the nearest whole number.%
Self-test 11- Equilibration of 2-butene at 327oC produces a mixture of Z-2-butene and E-2-butene that contains 60% of the E-isomer. What is the value of Keq ? Enter your answer to one decimal place.%
Example 2- Assume that Keq = 1 for the reaction . Table 1 summarizes the amounts of NaCl, KI, NaI, and KCl present at various stages of the reaction, starting with 1000 molecules of NaCl, 1000 molecules of KI and 0 molecules of NaI and KCl. The stoichiometry of the balanced equation indicates that one NaI and one KCl are formed for each NaCl and KI that react. Note that a value of Keq = 1 means that the ratio of products to reactants is 1/1.Since 1 + 1 = 2, at equilibrium 1/2 of all the molecules in the mixture will be products and 1/2 will be reactants.
Self-test 12- Assume that Keq = 2.0 for the reaction . Fill in the information that is missing in Table 2. Enter you answers in the same format as that given in each column of the table.
Self-test 13- Assume that Keq = 3 for the reaction . If you start with 1200 molecules of NaCl and 1200 molecules of KI, how many molecules of NaI will there be when the system comes to equilibrium?
PO6- You should be able to apply LeChâtelier's principle to determine how changing the concentration of one of the reactants or products in a system that is at equilibrium will change the concentration of the other reactants and products when the system returns to equilibrium.
Example 1- At 727oC Keq = 1 for the isomerization of Z-2-butene and E-2-butene. Imagine that you start with 1000 molecules of Z-2-butene and 0 molecules of the E-isomer. After the system comes to equilibrium, you perturb it by adding 300 more molecules of Z-2-butene. After the system returns to equilibrium, how many molecules of each compound will there be in the mixture? Since Keq = 1, the initial equilibrium mixture will contain 500 molecules of each of the two compounds involved in the reaction. Adding 300 molecules of Z-2-butene will momentarily increase the number of molecules of that isomer to 800. At this point Q = [500]/[800]= 0.625. Since this number is less than the equilibrium value of 1.00, this means that some of the added Z-2-butene will react to produce more of the E-isomer. Again, the stoichiometry of the balanced equation indicates that one E-2-butene is formed for each Z-2-butene that reacts. If n is the number of Z-2-butene molecules that react, then the expression for the equilibrium constant once equilibrium is reestablished is [500 + n]/[800 - n] = 1.00. Multiplying both sides by [800 - n] gives [500 + n] = [800 - n] which rearranges to yield 2n = 300. Therefore n = 150. Once the equilibrium is reestablished, the mixture will contain 650 molecules of E-2-butene and 650 molecules of Z-2-butene. These changes are summarized in Table 3.
Self-test 14- At 427oC Keq = 1.36 for the isomerization of Z-2-butene and E-2-butene. Imagine that you start with 1000 molecules of Z-2-butene and 0 molecules of the E-isomer. After the system comes to equilibrium, you perturb it by adding 300 more molecules of Z-2-butene. Select the number of molecules of Z-2-butene that will be present after each equilibrium is established.
500, then 551424, then 574 424, then 551 424, then 749
Self-test 15- Assume that Keq = 1 for the reaction . How will [NaI] change if you add more NaCl to the system?
The value of [NaI] will increase.The value of [NaI] will decrease. The value of [NaI] will not change. Not enough information is given. You need to know the concentrations of all the reactants and products.
Self-test 16- Consider the following reaction between calcium carbonate and sulfuric acid:. Calcium carbonate is a component in marble statues, while sulfuric acid is the principal component in acid rain. In other words, this is a reaction that describes how acid rain degrades statues in outdoor spaces. The CO2 that is formed in this reaction escapes as a gas. How does the loss of CO2 affect the reaction between the calcium carbonate and sulfuric acid?
It causes further reaction.It limits further reaction. It does not have any effect.It increases the amount of calcium carbonate, while decreasing the amount of sulfuric acid.
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