Thermochemistry

1. Thermochemistry By Sumana Ramakrishnan And Reema Sil

2. Thermochemistryis the study of heat change in chemical reactions. Endothermic is any process in which heat has to be supplied to the system from the surroundings. Exothermic is any process that gives off heat – transfers thermal energy from the system to the surroundings.

3. Enthalpyis used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. Hproducts < Hreactants Hproducts > Hreactants ∆H < 0 ∆H > 0 Exothermic (system gives off heat) Endothermic (system absorbs heat) Heat (q) absorbed or released: m : mass of a given quantity in grams (g) c : specific heat; the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius (Jg-1°C-1) ∆t = temperaturefinal – temperatureinitial (°C)

4. C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(l) When a 2.000-gram sample of pure phenol, C6H5OH(s), is completely burned according to the equation above, 64.98 kilojoules of heat is released. Use the information in the table below to answer the questions that follow. • Calculate the molar heat of combustion of phenol in kilojoules per mole at 25˚C. • Calculate the standard heat of formation, ∆H˚, of phenol in kilojoules per mole at 25˚C. • Calculate the value of the standard free-energy change, ∆G˚, for the combustion of phenol at 25˚C. • If the volume of the combustion container is 10.0 liters, calculate the final pressure in the container when the temperature is changed to 110˚C. (Assume no oxygen remains unreacted and that all products are gaseous.)

5. C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(l)When a 2.000-gram sample of pure phenol, C6H5OH(s), is completely burned according to the equation above, 64.98 kilojoules of heat is released. • Calculate the molar heat of combustion of phenol in kilojoules per mole at 25˚C. The molar heat of combustion of phenol is -3058 kilojoules.

6. C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(l)When a 2.000-gram sample of pure phenol, C6H5OH(s), is completely burned according to the equation above, 64.98 kilojoules of heat is released. Use the information in the table below to answer the questions that follow. (From (a), ∆Hrxn = -3058 kJ) (b) Calculate the standard heat of formation, ∆H˚, of phenol in kilojoules per mole at 25˚C. Equation Variables Numbers

7. C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(l)When a 2.000-gram sample of pure phenol, C6H5OH(s), is completely burned according to the equation above, 64.98 kilojoules of heat is released. Use the information in the table below to answer the questions that follow. (c) Calculate the value of the standard free-energy change, ∆G˚, for the combustion of phenol at 25˚C. First, you must solve for ∆S Then plug values into

8. C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(l)When a 2.000-gram sample of pure phenol, C6H5OH(s), is completely burned according to the equation above, 64.98 kilojoules of heat is released. Use the information in the table below to answer the questions that follow. (d) If the volume of the combustion container is 10.0 liters, calculate the final pressure in the container when the temperature is changed to 110˚C. (Assume no oxygen remains unreacted and that all products are gaseous.) } 0.1907 moles of gaseous products

9. 2 NO(g) + O2(g)→ 2 NO2(g) ∆H˚= -114.1 kJ, ∆S˚= -146.5 J K-1The reaction represented above is one that contributes significantly to the formation of photochemical smog. (a) Calculate the quantity of heat released when 73.1 g of NO(g) is converted to NO2(g). (b) For the reaction at 25˚C, the value of the standard free-energy change, ∆G˚, is -70.4 kJ. (i) Calculate the value of the equilibrium constant, Keq, for the reaction at 25˚C. (ii) Indicate whether the value of ∆G˚ would become more negative, less negative, or remain unchanged as the temperature is increased. Justify your answer. (c) Use the data in the table to calculate the value of the standard molar entropy, S˚, for O2(g) at 25˚C. (d) Use the data in the table to calculate the bond energy, in kJ mol-1, of the nitrogen-oxygen bond in NO2. Assume that the bonds in the NO2 molecule are equivalent (i.e., they have the same energy).

10. 2 NO(g) + O2(g)→ 2 NO2(g) ∆H˚= -114.1 kJ, ∆S˚= -146.5 J K-1The reaction represented above is one that contributes significantly to the formation of photochemical smog. (a) Calculate the quantity of heat released when 73.1 g of NO(g) is converted to NO2(g). The reason our answer is negative is because the reaction is exothermic. Even though it results in a negative change in heat for the reaction, the amount of heat released is 138.97kJ (a positive quantity). So the answer to the question in part (a) is the positive quantity 138.97 kJ.

11. 2 NO(g) + O2(g)→ 2 NO2(g) ∆H˚= -114.1 kJ, ∆S˚= -146.5 J K-1The reaction represented above is one that contributes significantly to the formation of photochemical smog. (b) For the reaction at 25˚C, the value of the standard free-energy change, ∆G˚, is -70.4 kJ. (i) Calculate the value of the equilibrium constant, Keq, for the reaction at 25˚C.

12. 2 NO(g) + O2(g)→ 2 NO2(g) ∆H˚= -114.1 kJ, ∆S˚= -146.5 J K-1The reaction represented above is one that contributes significantly to the formation of photochemical smog. (b) For the reaction at 25˚C, the value of the standard free-energy change, ∆G˚, is -70.4 kJ. (ii) Indicate whether the value of ∆G˚ would become more negative, less negative, or remain unchanged as the temperature is increased. Justify your answer. As T increases, the quantity T∆S becomes more and more negative. According to the equation, subtracting a large negative number (T∆S) from another negative number (∆H) is the same as adding a value to the negative ∆H, making ∆G˚ less negative. ∆G˚ will become less negative because of the roles of ∆H, T, and ∆S in the equation.

13. 2 NO(g) + O2(g)→ 2 NO2(g) ∆H˚= -114.1 kJ, ∆S˚= -146.5 J K-1The reaction represented above is one that contributes significantly to the formation of photochemical smog. (c) Use the data in the table below to calculate the value of the standard molar entropy, S˚, for O2(g) at 25˚C. ∆S˚ = ∑ S˚products - ∑ S˚reactants The standard molar entropy for O2(g)is 205.1 J K-1 mol-1.

14. 2 NO(g) + O2(g)→ 2 NO2(g) ∆H˚= -114.1 kJ, ∆S˚= -146.5 J K-1The reaction represented above is one that contributes significantly to the formation of photochemical smog. (d) Use the data in the table below to calculate the bond energy, in kJ mol-1, of the nitrogen-oxygen bond in NO2. Assume that the bonds in the NO2 molecule are equivalent (i.e., they have the same energy). (bonds of reactants are broken, bonds of products are formed) → In this equation, x represents the total bond energy in an NO2 molecule. Since there are two N-O bonds in NO2, x divided by two equals the bond energy of each N-O bond in the molecule.

15. Answer the following questions that relate to the chemistry of nitrogen. (a) Two nitrogen atoms combine to form a nitrogen molecule, as represented by the below equation. Using the table of average bond energies, determine the enthalpy change, ∆H, for the reaction. (b) The reaction between nitrogen and hydrogen to from ammonia is represented below. Predict the sign of the standard entropy change, ∆S˚, for the reaction. Justify your answer. ∆H˚= -92.2 kJ (c) The value of ∆G˚ for the reaction represented in part (b) is negative at low temperatures but positive at high temperatures. Explain. (d) When N2(g)and H2(g) are placed in a sealed contained at a low temperature, no measurable amount of NH3(g) is produced. Explain.

16. (a) Two nitrogen atoms combine to form a nitrogen molecule, as represented by the below equation. Using the table of average bond energies, determine the enthalpy change, ∆H, for the reaction. The enthalpy change when two nitrogen atoms combine is -950 kJ. The reaction only forms bonds, which requires heat energy. This makes the equation endothermic because it requires a heat supply to form N2(g).

17. (b) The reaction between nitrogen and hydrogen to from ammonia is represented below. Predict the sign of the standard entropy change, ∆S˚, for the reaction. Justify your answer. ∆S˚ (change in entropy) is the measure of disorder in a reaction. This measure of entropy can be judged by the change in moles of gaseous particles. In this reaction, since there are fewer gaseous moles of products than there are gaseous moles of reactants, there is a decrease in disorder - meaning, ∆S is negative for this reaction. ∆H˚= -92.2 kJ

18. (c) The value of ∆G˚ for the reaction represented in part (b) is negative at low temperatures but positive at high temperatures. Explain. The value of ∆G, free energy, is determined by the value of ∆H – T(∆S). We determined that both ∆H and ∆S are negative (from part b), and T will always be positive (because temperature in Kelvin is never negative. If T is a small value, T(∆S) will not be large enough to overcome ∆H. Thus, ∆G will be negative. If T is a large value, then T(∆S) will be large enough to overcome ∆H and make ∆G positive. So, subtracting the negative value of T(∆S) from ∆H can result in either a negative or positive number.

19. (d) When N2(g) and H2(g) are placed in a sealed contained at a low temperature, no measurable amount of NH3(g) is produced. Explain. As we determined in Part C, when at a low temperature, ∆G is negative. When ∆G is negative, it means that the reaction is spontaneous, proceeding in the forward direction on its own. However, if the reaction does not seem to visibly proceed, it does not mean that the reaction is not spontaneous. There are two reasons for why a measurable amount of NH3(g) is not observed. - All reactions require a certain activation energy to initiate the reaction. If a reaction does not have that required energy, the reaction will remain inert, even if it is spontaneous. - The reaction might require an extensive amount of time to proceed. Because the reaction is so slow, the amount of NH3(g) produced may not be immediately visible, but the reaction continues to occur.

20. A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the reaction represented above. The student combines equal volumes of 1.0M HCl and 1.0 M NaOH in an open polystyrene cup calorimeter. The heat released by the reaction is determined by using the equation q=mc ∆T.Assume the following. Both solutions are at the same temperature before they are combined. The densities of all the solutions are the same that of water. Any heat lost to the calorimeter of to the air is negligible. The specific heat capacity of the combined solutions is the same as that of water. (a) Give appropriate units for each of the terms in the equation q = mc ∆T. (b) List the measurements that must be made in order to obtain the value of q. (c) Explain how to calculate each of the following. I. The number of moles of water formed during the experiment II. The value of the molar enthalpy of neutralization, ∆Hneut, for the reaction between HCl(aq) and NaOH(aq) (d) The student repeats the experiment with the same equal volumes as before, but this time uses 2.0 M HCl and 2.0M NaOH. I. Indicate whether the value of q increases, decreases, or stays the same when compared to the first experiment. Justify your prediction. II. Indicate whether the value of the molar enthalpy of neutralization, ∆Hneut, increases, decreases or stays the same when compared to the first experiment. Justify your prediction. (e) Suppose that a significant amount of heat were lost to the air during the experiment. What effect would that have on the calculated value of the molar enthalpy of neutralization, ∆Hneut? Justify your answer.

21. A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the reaction represented above. The student combines equal volumes of 1.0M HCl and 1.0 M NaOH in an open polystyrene cup calorimeter. The heat released by the reaction is determined by using the equation q=mc ∆T.Assume the following.Both solutions are at the same temperature before they are combined. The densities of all the solutions are the same that of water. Any heat lost to the calorimeter of to the air is negligible. The specific heat capacity of the combined solutions is the same as that of water. (a) Give appropriate units for each of the terms in the equation q = mc ∆T.

22. A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the reaction represented above. The student combines equal volumes of 1.0M HCl and 1.0 M NaOH in an open polystyrene cup calorimeter. The heat released by the reaction is determined by using the equation q=mc ∆T.Assume the following.Both solutions are at the same temperature before they are combined. The densities of all the solutions are the same that of water. Any heat lost to the calorimeter of to the air is negligible. The specific heat capacity of the combined solutions is the same as that of water. (b) List the measurements that must be made in order to obtain the value of q. • Volume or mass of the HCl and NaOH solutions • Initial temperature of HCl and NaOH before mixing • Final temperature of HCl and NaOH after mixing

23. A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the reaction represented above. The student combines equal volumes of 1.0M HCl and 1.0 M NaOH in an open polystyrene cup calorimeter. The heat released by the reaction is determined by using the equation q=mc ∆T.Assume the following. Both solutions are at the same temperature before they are combined. The densities of all the solutions are the same that of water. Any heat lost to the calorimeter of to the air is negligible. The specific heat capacity of the combined solutions is the same as that of water. (c) Explain how to calculate each of the following. I. The number of moles of water formed during the experiment - Or - You could also reason that both HCl and NaOH react in a 1:1 ratio to form the same number of moles of water.

24. A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the reaction represented above. The student combines equal volumes of 1.0M HCl and 1.0 M NaOH in an open polystyrene cup calorimeter. The heat released by the reaction is determined by using the equation q=mc ∆T.Assume the following. Both solutions are at the same temperature before they are combined. The densities of all the solutions are the same that of water. Any heat lost to the calorimeter of to the air is negligible. The specific heat capacity of the combined solutions is the same as that of water. (c) Explain how to calculate each of the following. II. The value of the molar enthalpy of neutralization, ∆Hneut, for the reaction between HCl(aq) and NaOH(aq) First, you must solve for the heat produced using q=cm ∆T. Then, dividing this value by the moles of H2O (from part I). This gives you the molar enthalpy of neutralization for the reaction.

25. A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the reaction represented above. The student combines equal volumes of 1.0M HCl and 1.0 M NaOH in an open polystyrene cup calorimeter. The heat released by the reaction is determined by using the equation q=mc ∆T.Assume the following. Both solutions are at the same temperature before they are combined. The densities of all the solutions are the same that of water. Any heat lost to the calorimeter of to the air is negligible. The specific heat capacity of the combined solutions is the same as that of water. (d) The student repeats the experiment with the same equal volumes as before, but this time uses 2.0 M HCl and 2.0M NaOH. I. Indicate whether the value of q increases, decreases, or stays the same when compared to the first experiment. Justify your prediction. You can reason that since more moles of HCl and NaOH are reacting, the final temperature of the mixture will be higher. So ∆T will be greater, which will increase q.

26. A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the reaction represented above. The student combines equal volumes of 1.0M HCl and 1.0 M NaOH in an open polystyrene cup calorimeter. The heat released by the reaction is determined by using the equation q=mc ∆T.Assume the following. Both solutions are at the same temperature before they are combined. The densities of all the solutions are the same that of water. Any heat lost to the calorimeter of to the air is negligible. The specific heat capacity of the combined solutions is the same as that of water. (d) The student repeats the experiment with the same equal volumes as before, but this time uses 2.0 M HCl and 2.0M NaOH.II. Indicate whether the value of the molar enthalpy of neutralization, ∆Hneut, increases, decreases or stays the same when compared to the first experiment. Justify your prediction. As done in part C II, the molar enthalpy is found by dividing q by the moles of water. In the reaction, both q and water increase proportionally by x. Since x drops out, you can see that the value of the molar enthalpy of neutralization remains the same as in the first experiment.

27. A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the reaction represented above. The student combines equal volumes of 1.0M HCl and 1.0 M NaOH in an open polystyrene cup calorimeter. The heat released by the reaction is determined by using the equation q=mc ∆T.Assume the following. Both solutions are at the same temperature before they are combined. The densities of all the solutions are the same that of water. Any heat lost to the calorimeter of to the air is negligible. The specific heat capacity of the combined solutions is the same as that of water. (e) Suppose that a significant amount of heat were lost to the air during the experiment. What effect would that have on the calculated value of the molar enthalpy of neutralization, ∆Hneut? Justify your answer. Heat lost to the air will decrease ∆T – which in turn equals a smaller q. Thus, when you divide q by the number of moles of H2O, ∆Hneut will be a smaller value.

28. The End Voices by Sumana Ramakrishnan and Reema Sil Powerpoint by Reema Sil AP Problems painstakingly solved by both. Special Thanks to: Mr. Gangluff Mr. Amendola Eric Liu Jack Wang