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Worked Example 7.1 Heat of Reaction from Bond Energies

Worked Example 7.1 Heat of Reaction from Bond Energies. Estimate the (in kcal/mol) for the reaction of hydrogen and oxygen to form water:. Analysis.

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Worked Example 7.1 Heat of Reaction from Bond Energies

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  1. Worked Example 7.1 Heat of Reaction from Bond Energies Estimate the (in kcal/mol) for the reaction of hydrogen and oxygen to form water: Analysis The individual bond energies from Table 7.1 can be used to calculate the total bond energies of reactants and products. can then be calculated as Ballpark Estimate The average H— H bond energy is ~100 kcal/mol and the bond energy is ~120 kcal/mol. Thus, the total energy needed to break reactant bonds is ~ (200 + 120) = 320 kcal/mol. The O—H bonds are ~110 kcal/mol, so the total energy released when product bonds are formed is ~ 440 kcal/mol. Based on these estimates, –120 kcal/mol. Solution = (2(H—H) + ( )) – (4(O—H) ) = (2(103 kcal/mol) + (119 kcal/mol)) – (4(112 kcal/mol) ) = –123 kcal/mol Ballpark Check Our estimate was -120 kcal/mol, within 3% of the calculated answer.

  2. Worked Example 7.2 Heat of Reaction: Moles Methane undergoes combustion with according to the following equation: How much heat (in kcal and kJ) is released during the combustion of 0.35 mol of methane? Analysis Since the value of for the reaction (213 kcal/mol) is negative, it indicates the amount of heat released when 1 mol of methane reacts with . We need to find the amount of heat released when an amount other than 1 mol reacts, using appropriate factor-label calculations to convert from our known or given units to kilocalories, and then to kilojoules. Ballpark Estimate Since 213 kcal is released for each mole of methane that reacts, 0.35 mol of methane should release about one-third of 213 kcal, or about 70 kcal. There are about 4 kJ per kcal, so 70 kcal is about 280 kJ. Solution To find the amount of heat released (in kilocalories) by combustion of 0.35 mol of methane, we use a conversion factor of kcal/mol, and then we can convert to kilojoules using a kJ/kcal conversion factor (see Section 1.13): The negative sign indicates that the 75 kcal (314 kJ) of heat is released. Ballpark Check The calculated answer is consistent with our estimate (70 kcal or 280 kJ).

  3. Worked Example 7.3 Heat of Reaction: Mass to Mole Conversion How much heat is released during the combustion of 7.50 g of methane (molar mass = 16.0 g/mol)? Analysis We can find the moles of methane involved in the reaction by using the molecular weight in a mass to mole conversion, and then use to find the heat released. Ballpark Estimate Since 1 mol of methane (molar mass = 16.0 g/mol) has a mass of 16.0 g, 7.50 g of methane is a little less than 0.5 mol. Thus, less than half of 213 kcal, or about 100 kcal (418 kJ), is released from combustion of 7.50 g. Solution Going from a given mass of methane to the amount of heat released in a reaction requires that we first find the number of moles of methane by including molar mass (in mol/g) in the calculation and then converting moles to kilocalories or kilojoules: or The negative sign indicates that the 99.8 kcal 1418 kJ2 of heat is released. Ballpark Check Our estimate was –100 kcal (–418 kJ)!

  4. Worked Example 7.4 Heat of Reaction: Mole Ratio Calculations How much heat is released in kcal and kJ when 2.50 mol of reacts completely with methane? Analysis Since the for the reaction is based on the combustion of 1 mol of methane, we will need to perform a mole ratio calculation. Ballpark Estimate The balanced equation shows that 213 kcal (891 kJ) is released for each 2 mol of oxygen that reacts. Thus, 2.50 mol of oxygen should release a bit more than 213 kcal, perhaps about 250 kcal (1050 kJ). Solution To find the amount of heat released by combustion of 2.50 mol of oxygen, we include in our calculation a mole ratio based on the balanced chemical equation: or The negative sign indicates that the 266 kcal (1110 kJ) of heat is released. Ballpark Check The calculated answer is close to our estimate (–250 kcal or –1050 kJ).

  5. Worked Example 7.5 Entropy Change of Processes Does entropy increase or decrease in the following processes? (a) Smoke from a cigarette disperses throughout a room rather than remaining in a cloud over the smoker’s head. (b) Water boils, changing from liquid to vapor. (c) A chemical reaction occurs: Analysis Entropy is a measure of molecular disorder. Entropy increases when the products are more disordered than the reactants; entropy decreases when the products are less disordered than the reactants. Solution (a) Entropy increases because smoke particles are more disordered when they are randomly distributed in the larger volume. (b) Entropy increases because molecules have more freedom and disorder in the gas phase than in the liquid phase. (c) Entropy decreases because 4 mol of reactant gas particles become 2 mol of product gas particles, with a consequent decrease in freedom and disorder.

  6. Worked Example 7.6 Spontaneity of Reactions: Enthalpy, Entropy, and Free Energy The industrial method for synthesizing hydrogen by reaction of carbon with water has = +31.3 kcal/mol (+131 kJ/mol) and = +32 cal/[mol • K](+134 J/[mol • K]). What is the value of (in kcal and kJ) for the reaction at 27°C (300 K)? Is the reaction spontaneous or nonspontaneous at this temperature? Analysis The reaction is endothermic ( positive) and does not favor spontaneity, whereas the indicates an increase in disorder ( positive), which does favor spontaneity. Calculate to determine spontaneity. Ballpark Estimate The unfavorable (+31.3 kcal/mol) is 1000 times greater than the favorable (+32 cal/mol • K2), so the reaction will be spontaneous ( negative) only when the temperature is high enough to make the term in the equation for larger than the term. This happens at T 1000 K. Since T = 300 K, expect to be positive and the reaction to be nonspontaneous. Solution Use the free-energy equation to determine the value of at this temperature. (Remember that has units of calories per mole-kelvin or joules per mole-kelvin, not kilocalories per mole-kelvin or kilojoules per mole-kelvin.). Ballpark Check Because is positive, the reaction is nonspontaneous at 300 K, consistent with our estimate.

  7. Worked Example 7.7 Energy of Reactions: Energy Diagrams Draw an energy diagram for a reaction that is very fast but has a small negative free-energy change. Analysis A very fast reaction has a small . A reaction with a small negative free-energy change is a favorable reaction with a small energy difference between starting materials and products. Solution

  8. Worked Example 7.8 Writing Equilibrium Equations The first step in the industrial synthesis of hydrogen is the reaction of steam with methane to give carbon monoxide and hydrogen. Write the equilibrium equation for the reaction. Analysis The equilibrium constant K is the number obtained by multiplying the equilibrium concentrations of the products (CO and ) and dividing by the equilibrium concentrations of the reactants ( and ), with the concentration of each substance raised to the power of its coefficient in the balanced equation. Solution

  9. Worked Example 7.9 Equilibrium Equations: Calculating K In the reaction of with , the concentrations of reactants and products were determined experimentally at equilibrium and found to be 7.2 mol/L for , 7.2 mol/L for , and 0.050 mol/L for . Write the equilibrium equation, and calculate the equilibrium constant for the reaction. Which reaction is favored, the forward one or the reverse one? Analysis All the coefficients in the balanced equation are 1, so the equilibrium constant equals the concentration of the product, , divided by the product of the concentrations of the two reactants, and . Insert the values given for each concentration, and calculate the value of K . Ballpark Estimate At equilibrium, the concentration of the reactants (7.2 mol/L for each reactant) is higher than the concentration of the product (0.05 mol/L), so we expect a value of K less than 1. Solution The value of K is less than 1, so the reverse reaction is favored. Note that units for K are omitted. Ballpark Estimate Our calculated value of K is just as we predicted: K 1.

  10. Worked Example 7.10 Le Châtelier’s Principle and Equilibrium Mixtures Nitrogen reacts with oxygen to give NO: Explain the effects of the following changes on reactant and product concentrations: (a) Increasing temperature (b) Increasing the concentration of NO (c) Adding a catalyst Solution (a) The reaction is endothermic (positive ), so increasing the temperature favors the forward reaction. The concentration of NO will be higher at equilibrium. (b) Increasing the concentration of NO, a product, favors the reverse reaction. At equilibrium, the concentrations of both and , as well as that of NO, will be higher. (c) A catalyst accelerates the rate at which equilibrium is reached, but the concentrations at equilibrium do not change.

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