Determination of the Heat Capacity of a Calorimeter PDF

## Determination of the heat capacity of a calorimeter ### Basic concepts - Heat (q) is defined as the transfer of energy from or into a system because of a temperature difference between the system and its surroundings. Heat is often considered, inaccurately, as a form of energy existing as the r...

## Determination of the heat capacity of a calorimeter ### Basic concepts - Heat (q) is defined as the transfer of energy from or into a system because of a temperature difference between the system and its surroundings. Heat is often considered, inaccurately, as a form of energy existing as the result of the random motion of molecules. - Heat capacity (C) is defined as the amount of heat required to raise the temperature of a given quantity of the substance by one (Kelvin) degree. The units of heat capacity are J/°C. - Specific heat (Cs) is defined as the amount of heat required to raise the temperature of one gram of a substance by one (Kelvin) degree. The units of specific heat are J/gx°C or J/gxK since 1 °C = 1 K degree. ### Introduction Chemical reactions involving the transfer of heat are carried out in devices called calorimeters. Calorimeters are insulated to prevent loss or gain of heat between the calorimeter and its surroundings so that heat flow in the system can be measured. The total amount of heat that is produced or absorbed by a chemical reaction, at constant pressure, is called the enthalpy of reaction (ΔΗ). - If the chemical reaction is exothermic, heat is released, and the temperature of the system will be increased. * reactants → products + heat ΔΗ (-) - exothermic - If the chemical reaction is endothermic, heat is absorbed, and the temperature of the system will be decreased. * reactants + heat → products ΔΗ (+) - endothermic The quantity of heat absorbed or released (q) is calculated from the following equation: q = m x Cs x (Tf - Ti) Where: - q = amount of heat, measured in joules(J). - m = mass, measured in grams. - Cs = the specific heat. - Tf = final temperature. - Ti = initial temperature. ### Procedures - Put 50 mL of cold water in a clean and dry calorimeter and record the temperature every 30 s until the temperature becomes stable over 2 minutes and consider it as T1. - Prepare 50 mL of hot water and record its temperature and consider it as T2. - Add hot water to the calorimeter and record the temperature every 30 s until the temperature becomes stable over 2 minutes and consider it as T3. - Then the temperature starts to decrease, recode the first drop in temperature and consider it as T4. ### Calculation After equal quantities of hot and cold water are mixed in the calorimeter, the final equilibrium temperature is measured. The hot water loses heat to both the cold water and the calorimeter. In other words, heat gained by the cold water should equal that lost by the hot water. Any difference is due to heat gained by the calorimeter. This assumes no heat is lost to the surroundings from the calorimeter. Heat lost by hot water = heat gained by cold water + heat gained by the calorimeter. - qhot = qcold + qcal The minus sign indicates that the hot water is losing heat, whereas the cold water and the calorimeter are gaining heat. - - m. Cs. (T3 – T2) = m. Cs. (T3 - T1) + C (T4 - T1) ### Determination of the heat of solution ### Basic concepts - Heat of solution for saturated solution is defined as the change in heat content when one mole of solute is dissolved in a volume of water (solvent) to form a saturated solution. - Heat of solution for infinite dilution is defined as the change in heat content when one mole of solute is dissolved in a large amount of water (solvent) to form an infinitely dilute solution. ### Introduction Enthalpy of solution, or heat of solution, is expressed in kJ/mol, and it is the amount of heat energy that is released or absorbed when a solution is formed. There are three steps in the solvation process: 1. The breaking of bonds between solute molecules. 2. The breaking of intermolecular attractions between solvent molecules. 3. Formation of new solute-solvent attractive bonds. | Step | Energy change | | ------------------------ | ---------------------------------- | | Breaking solute bonds | Require and absorb energy | | Breaking solvent bonds | Require and absorb energy | | Forming solute-solvent bonds | release energy | Energy is absorbed during the first two steps, and it is released during the last step. Depending on the relative amounts of energy required to break bonds initially, as well as how much is released upon solute-solvent bond formation, the overall heat of the solution can either be endothermic or exothermic. - If the energy released is more than that absorbed, this process is exothermic, and the temperature of the system will be increased. * Solute + solvent → solution + heat ΔΗ (-) - exothermic - If the energy released is less than that absorbed, this process is endothermic, and the temperature of the system will be decreased. * Solute + solvent → solution + heat ΔΗ (+) - endothermic ### Procedures - Put 100 mL of water in a clean and dry calorimeter and record the temperature every 30 s until the temperature becomes stable over 2 minutes and consider it as T1. - Prepare 5 gm of solute. - Add the whole amount of solute (5 gm) to the calorimeter and record the temperature every 30 s until the temperature becomes stable over 2 minutes and consider it as T2. - Then the temperature starts to decrease, recode the first drop in temperature and consider it as T3. ### Calculation 1. Calculate the heat capacity of the calorimeter (C) [ from experiment no. 2]. 2. Calculate the number of moles present in 5 gm of solute. 3. Calculate the mass of 100 ml of water (m). 4. Search about the specific heat of water (Cs). 5. The heat of the solution is lost or gained to both the water and the calorimeter. In other words, heat gained should equal that loss. * Heat of solution = heat gained/lost by cold water + heat gained/lost by the calorimeter. - qsolution = qwater + qcal 6. Substitute all parameters in the previous equation and calculate the q solution 7. Calculate ΔΗ solution using the following equation * ΔΗ solution = q solution / number of moles of solute 8. Compare your result with the theoretical value and calculate the percent error. ## Determination of the heat of neutralization ### Basic concepts Heat of neutralization is defined as the amount of heat evolved when one equivalent of a strong acid and one equivalent of a strong base undergo a neutralization reaction to form water and salt. ### Introduction Strong acid reacts with a strong base and releases 13.7 kcal/mol - HCl + NaOH → NaCl + H2O ΔΗ =- 13.7 kcal/mol - HCl + KOH → KCl + H2O ΔΗ = - 13.7 kcal/mol **Why do strong acids reacting with strong alkalis give closely similar values?** We assume that strong acids and strong alkalis are fully ionized in solution and that the ions behave independently of each other. For example, dilute hydrochloric acid contains hydrogen ions and chloride ions in solution. Sodium hydroxide solution consists of sodium ions and hydroxide ions in solution. They react according to the following equation: Na+ (aq) + OH (aq) + H+ (aq) + Cl (aq) → Na+ (aq) + Cl (aq) + H2O(l) The equation for any strong acid being neutralized by a strong alkali is essentially just a reaction between hydrogen ions and hydroxide ions to make water. The other ions present (sodium and chloride, for example) are just spectator ions, taking no part in the reaction. The actual reaction between hydrochloric acid and sodium hydroxide solution is: H+ (aq) + OH (aq) → H2O(l) ΔΗ = - 13.7 kcal/mol ### Procedures - Put 50 mL of 2.5 M HCl in a clean and dry calorimeter and record the temperature every 30 s until the temperature becomes stable over 2 minutes and consider it as T1. - Put 50 mL of 2.5 M NaOH in a clean and dry beaker and record the temperature every 30 s until the temperature becomes stable over 2 minutes and consider it as T1. - Add the whole amount of NaOH to the calorimeter, and record the temperature every 30 s until the temperature becomes stable over 2 minutes and consider it as T2. - Then the temperature starts to decrease, recode the first drop in temperature and consider it as T3. ### Calculation 1. Calculate the heat capacity of the calorimeter (C) [ from experiment no. 2]. 2. Calculate the number of moles present in 50 ml of 2.5 M NaOH or HCl. * Number of moles of a substance in form of solution = concentration (mole/L) X volume (L) 3. Calculate the mass of 100 ml of NaCl solution (m). * 50 ml HCl + 50 ml NaOH = 100 ml NaCl solution 4. Search about the specific heat of NaCl solution (Cs). 5. The heat of the neutralization is lost to both the NaCl solution and the calorimeter. In other words, heat gained should equal that loss. * Heat of neutralization = [heat gained NaCl solution + heat gained the calorimeter]. - q neutralization = qNaCl sol + qcal 6. Substitute all parameters in the previous equation and calculate the q neut 7. Calculate ΔΗneut using the following equation * ΔΗneut = q neutralization / number of moles of HCl or NaOH 8. Compare your result with the theoretical value and calculate the percent error. ## Determination of the heat of displacement ### Basic concepts Heat of displacement is defined as the heat change when one mole of metal is displaced from its salt solution by a more reactive metal. ### Introduction The more reactive (electropositive) metal can able to displaces a less reactive metal from a solution of its salt. - Zn + CuSO4 → ZnSO4 + Cu This reaction is an oxidation-reduction reaction as the following - Zn → Zn+2 + 2e - oxidation process - Cu+2 + 2e → Cu - reduction process - Zn + Cu+2 → Zn+2 + Cu - overall reaction As you know the zinc metal can displace the copper from the CuSO4 solution, while the reverse reaction is not possible. These phenomena due to the fact of electropositivity. Electropositivity is the ability of metals to lose electrons (oxidation process). The more electropositive metal has a greater tendency to lose electrons than the less electropositive metal. Therefore the zinc metal can displace copper while the reverse reaction is not possible. **What if we use Mg metal instead of Zn metal. Is it able to replace the copper? Why?** **Why you should know about the Electrochemical series to be able to answer the previous questions?** ### Procedures - Put 50 mL of 0.25 M CuSO4 in a clean and dry calorimeter and record the temperature every 30 s until the temperature becomes stable over 2 minutes and consider it as T1. - Prepare 1 gm of zinc metal powder. - Add the whole amount of zinc metal powder to the calorimeter, and record the temperature every 30 s until the temperature becomes stable over 2 minutes and consider it as T2. - Then the temperature starts to decrease, recode the first drop in temperature and consider it as T3. ### Calculation 1. Calculate the heat capacity of the calorimeter (C) [ from experiment no. 2]. 2. Calculate the number of moles present in 50 ml of 0.25 M CuSO4 and 1 gm of Zn. * Number of moles of a substance in the solution form = concentration (mole/L) x volume (L) * Number of moles of a substance in the solid form = mass (g) / M.Wt or A.Wt (g/mole) 3. Calculate the mass of 50 ml of ZnSO4 solution (m). 4. Search about the specific heat of ZnSO4 solution (Cs). 5. The heat of the displacement is lost to both the ZnSO4 solution and the calorimeter. In other words, heat gained should equal that loss. * Heat of displacement = [heat gained ZnSO4 solution + heat gained the calorimeter]. - q displacement = q ZnSO4 sol + qcal 6. Substitute all parameters in the previous equation and calculate the q displacement 7. Search about the limiting agent. 8. Calculate ΔΗdis using the following equation * ΔΗdis = q displacement / number of moles of limiting agent 9. Compare your result with the theoretical value and calculate the percent error. ## Determination of the specific heat of metals ### Basic concepts - Specific heat (Cs) is defined as the amount of heat required to raise the temperature of one gram of a substance by one (Kelvin) degree. - The units of specific heat are J/gx°C or J/gxK since 1 °C = 1 K degree. ### Purpose The purpose of this lab is to apply the experimental methods of calorimetry in the determination of the specific heat of several metals. ### Procedures - Put 50 mL of cold water in a clean and dry calorimeter and record the temperature every 30 s until the temperature becomes stable over 2 minutes and consider it as T1. - Prepare a piece of iron metal and polish it using emery then clean and dry it. - Weigh the iron piece using a two-digit balance. - Put the iron metal piece in a water path and thermoset at a temperature near to the boiling point of water, until the temperature becomes stable over 2 minutes, and consider it as T2. - Quickly insert the hot iron piece into the calorimeter and record the temperature every 30 s until the temperature becomes stable over 2 minutes and consider it as T3. - Then the temperature starts to decrease, recode the first drop in temperature and consider it as T4. ### Calculation 1. Calculate the heat capacity of the calorimeter (C) [ from experiment no. 2]. 2. Search about the specific heat of water (Cs). 3. The iron piece loses heat to both the cold water and the calorimeter. In other words, heat gained should equal that loss. - Heat lost by iron piece = heat gained by cold water + heat gained by the calorimeter - q iron piece = q water + qcal 4. Substitute all parameters in the previous equation and calculate the specific heat of iron (Cs) 5. Compare your result with the theoretical value and calculate the percent error. 6. Repeat the experiment with other types of metals