Chem 317 (for Chem 305) Exercise # 10 Energy, Rates, and Equilibrium of Reactions

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1 Exercise # 10 Energy, Rates, and Equilibrium of Reactions PART 1: ENERGY OF REACTIONS Enthalpy: During a reaction, heat can be given off or consumed during the making and breaking of bonds. The heat change during a reaction at constant pressure is known as the enthalpy change, given the symbol H. Exercise #1. Answer the following a True (T) or False (F). If heat is given off during a reaction: T Heat is a product T The products are lower in energy than the reactants. F The sign of H is positive. F The reaction is said to be endergonic. Heat of reaction: The enthalpy of a reaction is an extensive property. This means that it is dependant on the amount of material that is put into the reaction. The more moles or grams of material used in the reaction, the more heat will be produced (if it is an exothermic reaction) or consumed (if it is an endothermic reaction). Exercise #. How much heat is released or consumed during the combustion of.75g of methane (CH ), given the following: CH (g) + O (g) CO (g) + H O(l) H = -1kcal/mol CH Hint: Set this up as a dimensional analysis problem!.75g CH 1mol CH 16.0g CH 1kcal 1mol CH = 9.8kcal Entropy: Entropy (S) is a measure of the randomness and dispersion of matter. Since the universe does not tend to favor processes that increase organization, the chances of a reaction happening spontaneously increase if the change in entropy for the process ( S) is positive. Exercise #. Answer the following a True (T) or False (F). ( S) is positive for the following processes: T ice melting T helium escaping from a balloon F steam condensing. F ions forming a precipitate Free Energy: A reaction is more likely to be spontaneous if it is exothermic (- H). It is also more likely to be spontaneous if it has an increase in entropy (+ S). These two factors must be looked at together to determine if the reaction spontaneous or not. Reaction spontaneity is determined by the Partially adapted from The Chem Team 1

2 Gibbs Free Energy ( G). This is calculated using the formula: G = H - T S The reaction will be spontaneous (exergonic) if G is (-). The reaction will be non-spontaneous (endergonic) if G is (+) Note: T must be in Kelvin H and S must both be in J kj cal, or kcal! (You will need to convert one since they are usually not expressed in the same units! Exercise #. Is the following reaction spontaneous or non-spontaneous at 5 C if H = -91.8kJ and S=-198.0J/K at this temperature? G = H T S = 91.8kJ N (g) + H (g) NH (g) G = -91.8kJ [(98.15K)( kJ/K) G = -91.8kJ (-59.0kJ) = -.8kJ PART : RATES OF REACTIONS 198.0J kj [( ) K] K 1000J The rate of a reaction is influenced by a variety of factors. Although we will not calculate the actual mathematical relationships, it is important to understand the way in which these factors influence reaction rates. Exercise #5. Complete the following. 1. Which of the following does NOT influence the rate of reaction? a) temperature b) concentration c) particle size d) none of these answers. When the temperature goes up, the reaction rate goes up.. When the concentration of reactants goes up, the reaction rate goes up.. When the particle size goes up, the reaction rate goes down. 5. When a catalyst is added, the reaction rate goes up. 6. The energy hurdle that reactants must go over to become products is known as the activation energy. Partially adapted from The Chem Team

3 7. a) Please identify the letter code for Ea and H on the following diagrams. b) Also, identify them as exothermic or endothermic reactions. Ea = a H = c exothermic or endothermic? endo Ea = a H = b exo PART REACTION EQUILIBRIUM When a system is at uilibrium, a constant value is established by the multiplicative product of the concentrations of the products' concentrations (each raised first to the power of its coefficient), then divided by the multiplicative product of the reactants' concentrations (each raised first to the power of its coefficient). Now that may be somewhat wordy, so here it is using a generic chemical uation: wa + xb yc + zd Following the word definition above, we have this: A, B, C, and D are the chemical substances. x, y, w and z are the coefficients. Let s do it again, this time using a specific chemical uation: And the answer is: SO + O SO Partially adapted from The Chem Team

4 Exercise #6. Write an uilibrium expression for each of the following reactions. 1) O <===> O ) PCl 5 <===> PCl + Cl ) N + H <===> NH 5) SO + (1/) O <===> SO ) H + I <===> HI Calculating the Equilibrium Constant from Equilibrium Conditions The easiest way to explain is by an example problem or two: 1) Calculate the uilibrium constant (K ) for the following reaction: H + I <==> HI when the uilibrium concentrations at 5 C were found to be: [H ] = M [I ] = M [HI] = 0.89 M The first thing to do is write the uilibrium expression for the reaction as written. Now, all you have to do is substitute numbers into the problem. K is what we want to find, so that's our "x." Here is what we get: Exercise #7.: x = (0.89) = 60. [(0.0505) (0.098)] Calculate K using the same uation as above and with the following uilibrium concentrations: [H ] = M [I ] = M [HI] = M K= (0.017) = 8. (0.0060)( ) Partially adapted from The Chem Team

5 Using a Known K If we know the value of the uilibrium constant, and the concentration of all reactants and products except one, we can use the uilibrium constant expression and a little algebra (I know that is a scary word!) to solve for the one we don t know. Example: For the reaction N + H NH If K = 6.0x10 -, Find the uilibrium concentration of N if the uilibrium concentrations are: [ NH ] - [NH ] = M [H ] = 0.76 M so K = = 6. 0 x 10 [ N ][H ] We need to rearrange the uation to solve for [N ]. We can do this by multiplying both sides of the uation by [N ] and dividing both sides by K. This results in: [ NH ] ( 0.157) [N ] = = = = 0.9 M - (K )[H ] (6.0 x 10 )(0.76) (0.060)(0.) Exercise #8. Given the uation SO + O SO K =.6 Calculate the uilibrium concentration of oxygen given the following uilibrium concentrations: [SO ] = 1.50 [SO ] =.50 [SO ] K = [SO ].50 [O ] [SO ] [O ] K [SO ] 1.5 M = = = (.6)(1.50) Using K to Predict Relative Concentrations The size of the uilibrium constant can give us information about the relative amounts of reactants and products present at uilibrium. When K << 1 The value in the numerator (representing product concentrations) is much smaller than the value in the denominator (representing reactant concentrations). The reaction lies to the left (mostly reactants) When K >> 1 The value in the denominator (representing reactant concentrations) is much smaller than the value in the numerator (representing product concentrations). The reaction lies to the right (mostly products) When K 1 The value in the denominator (representing reactant concentrations) is close to the value in the numerator (representing product concentrations). The reaction lies in the middle (mix of reactants and products) Partially adapted from The Chem Team 5

6 Exercise #9. Use K to predict the relative concentrations of reactants and products for the following uilibria. 1) HI + H O I - + H O + K =.5 x a) mostly products b) mostly reactants c) mix of reactants and products Ans: a) ) HF + H O F - + H O + K = 6.6 x 10 - a) mostly products b) mostly reactants c) mix of reactants and products Ans: b) ) Which acid HI or HF is a strong acid and why? The stronger the acid, the more it breaks up into its constituent ion. Since HI has more product ions than HF, HI is the stronger acid. Le Chatlier s Principle This principle states that if a system at uilibrium is disturbed, the uilibrium position will shift to relieve the stress put on the system. - Adding reactants (including heat) will shift a reaction to the right. - Adding products (including heat)* will shift a reaction to the left. - Removing (reacting away) a reactant will shift a reaction to the left. - Removing (reacting away) a product will shift a reaction to the right. - Changes to amounts of solids and liquids do not shift the position of the uilibrium because they do not appear in the uilibrium constant expression! Exercise #10. How will the following changes alter the uilibrium for the reaction: Fe (s) + H O (g) Fe O (s) + H (g) H o = -150 kj SHIFT a. H O is removed from the system. LEFT b. H is removed from the system. RIGHT c. Fe O is added to the system. No effect (it s a solid!) d. The temperature is raised. Left Partially adapted from The Chem Team 6

Energy Changes, Reaction Rates and Equilibrium. Thermodynamics: study of energy, work and heat. Kinetic energy: energy of motion

Energy Changes, Reaction Rates and Equilibrium Thermodynamics: study of energy, work and heat Kinetic energy: energy of motion Potential energy: energy of position, stored energy Chemical reactions involve