Calorimetry Lab
Calorimetry
To determine the specific heat of a metal and its approximate atomic mass. To determine the heat of neutralization for a strong acid-strong base reaction. To determine the quantity and direction of heat flow for the dissolution of salt. Post Lab Questions and Answers:
1. In parts A and B in, the calorimeter, although a good insulator, absorbs some heat when the system is above room temperature. Is the reported value for the specific heat of the metal too high or too low? Explain. Is the reported DeltaHn value for the acid- base reaction too high or too low? Explain. The DeltaHn value reported in Part B will be lower than the true value because the true Delta T will never be reached. Since the calorimeter absorbs some …show more content…
heat when the system is above room temperature the DeltaT will be too low. Since DeltaT is too low, q in q=mc DeltaT equation will be too low. Therefore, DeltaH is too low.
2. The DeltaHn value for the two strong acid-strong base reactions should be the same, within experimental error. Explain.
The DeltaHn values for the two strong acid-strong bases reactions should be the same, within experimental error, because the net ionic equations are the same for both reactions.
3. If you use a thermometer that is miscalibrated to read .4 degrees Celsius higher over its entire range, does this affect the value for DeltaHn? Explain. If a thermometer is miscalibrated to read .4 degrees Celsius higher over its entire range the DeltaHn value will be unaffected because the DeltaT will be unchanged due to the miscalibration.
4. If the maximum recorded temperature is used in Part B, rather that the extrapolated temperature, will be reported DeltaHn be higher or lower than the accepted value? Explain. If the maximum recorded temperature is used in Part B, rather that the extrapolated temperature, DeltaHn will be too small. This is due to the fact that the maximum temperature is lower than the temperature of the line of best fit after one minute. Conclusion: The purposes of this lab were achieved as we determined both the heat of neutralization for a strong acid-base reaction and the quantity and direction of heat flow for the dissolution of salt. Heat of neutralization is determined by assuming the density and the specific heats of the acids and the base solutions are equal to that of water, and measuring the temperature change when the two are mixed. In order to determine the heat of neutralization of the HCL and the NaOH mixture, we first took a volume of the acid (50.0 mL HCL) and found the temperature of the acid to be 22.2 using the thermometer placed in our calorimeter. A calorimeter is the laboratory apparatus that is used to measure the quantity and direction of heat flow accompanying a chemical or physical change. This heat flow for the chemical reactions is quantitatively expressed as the heat of reaction. We stirred the acid to develop a base line on our graph, and then added 50.0 mL of NaOH at 22.2 degrees Celsius with a given molarity of 1.038. Stirring the acid and base together for about eight minutes gave us a graph showing the temperature change over time between the two reactants. With this temperature over time graph we were able to extrapolate a maximum temperature for the reaction to be 29.5 degrees Celsius. The temperature change was 7.5 degrees Celsius. Using a simple equation, and the given density of the solution being 1.0 g/mol we found the mass of the final mixture to be 100 g. Using the equation q=mcDeltaT, and the previously determined mass in grams, the given specific heat of solution, and the change in temperature, we were able to find q, or the heat of solution to be 3100 J. The heat of the reaction is the opposite of the heat of solution so the desired reaction is negative 3100 J. Simple stoichiometry equations using the given volume and density of the NaOH yield the moles of OH^-. The heat evolved per mole of the water is found by dividing the approximately -3100 j by the .0519 moles of water formed resulting in – 60.5 kJ/mol. The same steps were followed for the reaction of HNO3 and NaOH, resulting in approximately – 3100 J evolved divided by the amount of moles of water resulting in a heat evolved per mole of water of -59.7 kJ/ mol. The expected values for both the reaction for the heat evolved per mole of water is given as -57.8 kJ/mol. We then determined the percent error. For the HCl it was 4.67 % and for the HNO3 it was 3.29% This error is due to the calorimeter absorbing heat during the reaction, and the loss of heat during incorrect movements of students, which results in a larger change in temperature which when put into a q=mcDeltaT equation produces a more negative number of joules, which when dividing by the amount of moles of water produces a value more negative (lower) than the expected value. Part C of the lab was to determine the heat of solution for the dissolving of a slat.
We were given KOH and KNO3 as our salts. We weighed the calorimeter and water then subtracted accordingly o get the mass of the calorimeter and that of the water. The lab instructed on the amounts of water and salt to be used. The initial temperature of the water in the KOH reaction was 22.8 degrees Celsius and the final temperature of the mixture determined from the graph was 37.0 degrees Celsius. The graph of the KOh showed it to be an exothermic salt and the KNO3 to be and endothermic salt because of the change in the temperature indicating either absorbing or giving off of heat by the salts. This the change in temperature of the solution of the KOH and water is + 28880 J and the heat change of the salt is 82.6 J. Thus the total heat change, in the reaction is -2960 J. To determine the DeltaHs (J/g salt), divide the totalt heat of change by the measure mass of the salt resulting in -590 J/g. The same process was followed for KNO3 but this solution had a -5.9 degree Celsius temperature change this producing a negative total heat change of water and the slat in the solution giving a +13000 J in the reaction. DeltaHs is then determines to be 250 J/g. The expected value of DeltaHs of KNO3 is +354 J/g and for KOH, -1026 J/g. The error for the KNO3 is -27.5% and for the KOH it is 42.5%. These errors are high but not uncommon for the experiment. They are caused because the salts are in a solid form and thus after the peak of temperature of solution is found, the remainder of the graph is flawed because salt is still in the process of dissolving. This dissolving still of the salts is giving off or absorbing heat thus making the extrapolated change in the temperature of solution to be small. The temperature of solution being too small results in the total heat change being too small, giving the DeltaHs value as in the case of KOH too high (not negative enough) and in the KNO3
being too low. Many class room topics where reinforced in this lab. Especially the use of the equation q=mcDeltaT, DeltaH values, Heat of solution values, specific heat, and percent error. We also reinforced the skills of measuring liquids, working with chemicals, and many more. This was the first time we worked with a calorimeter and the computer graphing. This was also the first time we explored lattice energy and hydration energy of a salt. Possible sources of error could be in spilling the liquids out of the cup, malfunction of the graphing utility, mixing the wrong substances, and not stirring the reaction. All of these would have skewed the data. I did not have any predicted results because I was not present for the first day of the lab. Therefore my results cannot be compared.
To determine the specific heat of a metal and its approximate atomic mass. To determine the heat of neutralization for a strong acid-strong base reaction. To determine the quantity and direction of heat flow for the dissolution of salt. Post Lab Questions and Answers:
1. In parts A and B in, the calorimeter, although a good insulator, absorbs some heat when the system is above room temperature. Is the reported value for the specific heat of the metal too high or too low? Explain. Is the reported DeltaHn value for the acid- base reaction too high or too low? Explain. The DeltaHn value reported in Part B will be lower than the true value because the true Delta T will never be reached. Since the calorimeter absorbs some …show more content…
heat when the system is above room temperature the DeltaT will be too low. Since DeltaT is too low, q in q=mc DeltaT equation will be too low. Therefore, DeltaH is too low.
2. The DeltaHn value for the two strong acid-strong base reactions should be the same, within experimental error. Explain.
The DeltaHn values for the two strong acid-strong bases reactions should be the same, within experimental error, because the net ionic equations are the same for both reactions.
3. If you use a thermometer that is miscalibrated to read .4 degrees Celsius higher over its entire range, does this affect the value for DeltaHn? Explain. If a thermometer is miscalibrated to read .4 degrees Celsius higher over its entire range the DeltaHn value will be unaffected because the DeltaT will be unchanged due to the miscalibration.
4. If the maximum recorded temperature is used in Part B, rather that the extrapolated temperature, will be reported DeltaHn be higher or lower than the accepted value? Explain. If the maximum recorded temperature is used in Part B, rather that the extrapolated temperature, DeltaHn will be too small. This is due to the fact that the maximum temperature is lower than the temperature of the line of best fit after one minute. Conclusion: The purposes of this lab were achieved as we determined both the heat of neutralization for a strong acid-base reaction and the quantity and direction of heat flow for the dissolution of salt. Heat of neutralization is determined by assuming the density and the specific heats of the acids and the base solutions are equal to that of water, and measuring the temperature change when the two are mixed. In order to determine the heat of neutralization of the HCL and the NaOH mixture, we first took a volume of the acid (50.0 mL HCL) and found the temperature of the acid to be 22.2 using the thermometer placed in our calorimeter. A calorimeter is the laboratory apparatus that is used to measure the quantity and direction of heat flow accompanying a chemical or physical change. This heat flow for the chemical reactions is quantitatively expressed as the heat of reaction. We stirred the acid to develop a base line on our graph, and then added 50.0 mL of NaOH at 22.2 degrees Celsius with a given molarity of 1.038. Stirring the acid and base together for about eight minutes gave us a graph showing the temperature change over time between the two reactants. With this temperature over time graph we were able to extrapolate a maximum temperature for the reaction to be 29.5 degrees Celsius. The temperature change was 7.5 degrees Celsius. Using a simple equation, and the given density of the solution being 1.0 g/mol we found the mass of the final mixture to be 100 g. Using the equation q=mcDeltaT, and the previously determined mass in grams, the given specific heat of solution, and the change in temperature, we were able to find q, or the heat of solution to be 3100 J. The heat of the reaction is the opposite of the heat of solution so the desired reaction is negative 3100 J. Simple stoichiometry equations using the given volume and density of the NaOH yield the moles of OH^-. The heat evolved per mole of the water is found by dividing the approximately -3100 j by the .0519 moles of water formed resulting in – 60.5 kJ/mol. The same steps were followed for the reaction of HNO3 and NaOH, resulting in approximately – 3100 J evolved divided by the amount of moles of water resulting in a heat evolved per mole of water of -59.7 kJ/ mol. The expected values for both the reaction for the heat evolved per mole of water is given as -57.8 kJ/mol. We then determined the percent error. For the HCl it was 4.67 % and for the HNO3 it was 3.29% This error is due to the calorimeter absorbing heat during the reaction, and the loss of heat during incorrect movements of students, which results in a larger change in temperature which when put into a q=mcDeltaT equation produces a more negative number of joules, which when dividing by the amount of moles of water produces a value more negative (lower) than the expected value. Part C of the lab was to determine the heat of solution for the dissolving of a slat.
We were given KOH and KNO3 as our salts. We weighed the calorimeter and water then subtracted accordingly o get the mass of the calorimeter and that of the water. The lab instructed on the amounts of water and salt to be used. The initial temperature of the water in the KOH reaction was 22.8 degrees Celsius and the final temperature of the mixture determined from the graph was 37.0 degrees Celsius. The graph of the KOh showed it to be an exothermic salt and the KNO3 to be and endothermic salt because of the change in the temperature indicating either absorbing or giving off of heat by the salts. This the change in temperature of the solution of the KOH and water is + 28880 J and the heat change of the salt is 82.6 J. Thus the total heat change, in the reaction is -2960 J. To determine the DeltaHs (J/g salt), divide the totalt heat of change by the measure mass of the salt resulting in -590 J/g. The same process was followed for KNO3 but this solution had a -5.9 degree Celsius temperature change this producing a negative total heat change of water and the slat in the solution giving a +13000 J in the reaction. DeltaHs is then determines to be 250 J/g. The expected value of DeltaHs of KNO3 is +354 J/g and for KOH, -1026 J/g. The error for the KNO3 is -27.5% and for the KOH it is 42.5%. These errors are high but not uncommon for the experiment. They are caused because the salts are in a solid form and thus after the peak of temperature of solution is found, the remainder of the graph is flawed because salt is still in the process of dissolving. This dissolving still of the salts is giving off or absorbing heat thus making the extrapolated change in the temperature of solution to be small. The temperature of solution being too small results in the total heat change being too small, giving the DeltaHs value as in the case of KOH too high (not negative enough) and in the KNO3
being too low. Many class room topics where reinforced in this lab. Especially the use of the equation q=mcDeltaT, DeltaH values, Heat of solution values, specific heat, and percent error. We also reinforced the skills of measuring liquids, working with chemicals, and many more. This was the first time we worked with a calorimeter and the computer graphing. This was also the first time we explored lattice energy and hydration energy of a salt. Possible sources of error could be in spilling the liquids out of the cup, malfunction of the graphing utility, mixing the wrong substances, and not stirring the reaction. All of these would have skewed the data. I did not have any predicted results because I was not present for the first day of the lab. Therefore my results cannot be compared.