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molM₂ msoln Part I: Mg + HCI Moles Mg Mass Mg + HCI Soln (g)

molM₂ msoln Part I: Mg + HCI Moles Mg Mass Mg + HCI Soln (g)

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Can you double check my part one calculations... The first pic is the equations we are supposed to use. The second pic at the top is the data I collected during the trials for part one and 2. I know I got part 2 right I just need to know where I'm messing up on part one.
1. 2. 3. 4. 5. 6. 7. 8. 3. 4. 5. 9:28 6. Done 161_8_+Enthalpy_F2020 7. Part I: Mg + HCI Mass Mg (g) Mass HCI Soln (g) Initial Temp (°C) Final Temp (°C) molMg msoln AT qsoln 1. molmgo 1. Calculate Moles Mg from mass Mg 2. Add Mg and HCl(aq) mass together 3. AT=Tr-Ti 4. See Equation 3 of the Background Part I: Mg + HCI qcal 2. msoln qrxn ΔΗ1,2 AH AT qsoln qcal qran Trial 1 0.2536 99.9253 AH1.2 Trial 2 0.2532 100.1430 24.94 25 35.42 37.88 Moles Mg Mass Mg + HCI Soln (g) Change in Temp (°C) Heat Absorbed by Soln (J) [Ssotn = 4.184 Heat Absorbed by Cal (J) [Ccal = 20.1 Heat Released by Reaction (J) Enthalpy of Reaction (kJ) Avg Enthalpy of Reaction (kJ) Part II: MgO + HCI Moles MgO Mass MgO + HCI Soln (g) Change in Temp (°C) Heat Absorbed by Soln (J) [Ssoln 4.184- // Heat Absorbed by Cal (J) [Ccal = 20./1 Heat Released by Reaction (J) Enthalpy of Reaction (kJ) 8. AHxn Avg Enthalpy of Reaction (kJ) Part II: MgO + HCI Mass MgO (g) Mass HCI Soln (g) Initial Temp (°C) Final Temp (°C) Trial 1 5. See Equation 2 of the Background 6.-qxn [qsoln+qcal] 7. See Equation 4 of the Background Average AH for trials 1 and 2 8. Trial 2 0.01043 100.1789 10.58 4434.59 211.6 -4646.19 -445.46 Trial 1 0.0248 100.8596 4.83 2038.24 96.6 2134.84 1 -86.08 1 # Trial 1 Trial 2 0.9995 0.9990 99.8601 99.5041 26.07 30.77 25.97 30.80 -497.87 0.0103 100.3952 12.88 5410.28 257.6 -5667.88 -550.28 Trial 2 0.02478 84.81 100.4941 4.7 1976.20 94 -2070.2 -83.54 Show work for Part I Trial 1 for calculations 4, 5, 6, 7 (attach additional paper if necessary):
Transcribed Image Text:1. 2. 3. 4. 5. 6. 7. 8. 3. 4. 5. 9:28 6. Done 161_8_+Enthalpy_F2020 7. Part I: Mg + HCI Mass Mg (g) Mass HCI Soln (g) Initial Temp (°C) Final Temp (°C) molMg msoln AT qsoln 1. molmgo 1. Calculate Moles Mg from mass Mg 2. Add Mg and HCl(aq) mass together 3. AT=Tr-Ti 4. See Equation 3 of the Background Part I: Mg + HCI qcal 2. msoln qrxn ΔΗ1,2 AH AT qsoln qcal qran Trial 1 0.2536 99.9253 AH1.2 Trial 2 0.2532 100.1430 24.94 25 35.42 37.88 Moles Mg Mass Mg + HCI Soln (g) Change in Temp (°C) Heat Absorbed by Soln (J) [Ssotn = 4.184 Heat Absorbed by Cal (J) [Ccal = 20.1 Heat Released by Reaction (J) Enthalpy of Reaction (kJ) Avg Enthalpy of Reaction (kJ) Part II: MgO + HCI Moles MgO Mass MgO + HCI Soln (g) Change in Temp (°C) Heat Absorbed by Soln (J) [Ssoln 4.184- // Heat Absorbed by Cal (J) [Ccal = 20./1 Heat Released by Reaction (J) Enthalpy of Reaction (kJ) 8. AHxn Avg Enthalpy of Reaction (kJ) Part II: MgO + HCI Mass MgO (g) Mass HCI Soln (g) Initial Temp (°C) Final Temp (°C) Trial 1 5. See Equation 2 of the Background 6.-qxn [qsoln+qcal] 7. See Equation 4 of the Background Average AH for trials 1 and 2 8. Trial 2 0.01043 100.1789 10.58 4434.59 211.6 -4646.19 -445.46 Trial 1 0.0248 100.8596 4.83 2038.24 96.6 2134.84 1 -86.08 1 # Trial 1 Trial 2 0.9995 0.9990 99.8601 99.5041 26.07 30.77 25.97 30.80 -497.87 0.0103 100.3952 12.88 5410.28 257.6 -5667.88 -550.28 Trial 2 0.02478 84.81 100.4941 4.7 1976.20 94 -2070.2 -83.54 Show work for Part I Trial 1 for calculations 4, 5, 6, 7 (attach additional paper if necessary):
Figure 1: Coffee Cup Calorimeter -Greaction=+qsurroundings Equation 1: Calorimetry Assumption Example 1: Heat Capacity Ex. When any substance absorbs heat, it experiences a change in temperature, but every substance has a unique ability to "store" heat. This is called its heat capacity (C), defined as the amount of heat required to raise the temperature of that substance by 1°C. Thus, changing the temperature of the sample from an initial temperature (T) to a final temperature (T) requires heat equal to: q=C.AT Equation 2: Heat transfer as a function of heat capacity and temperature change heat (a), w Example 1: An object with a heat capacity of 20. J/°C has its temperature raised from 22.0°C to 33.5°C. How much heat energy would this require? q=C AT = 20 (33.5°C -22.0°C) = 230 J Reaction Surroundings Insulated Walls Figure 2: Simplified Calorimeter Heat capacity is often scaled to be measured per gram of a substance, a related form known as the specific heat capacity (S), defined as the amount of heat required to raise the temperature of 1 gram of a substance by 1°C. q=S·m AT= 4.184- 4T q=S.m.AT Equation 3: Heat transfer as a function of specific heat capacity, mass, and temperature change gºC Example 2: Specific Heat Capacity Ex. Example 2: A water solution with a specific heat capacity of 4.184 J/g°C and weighing 100. g has its temperature raised from 22.0°C to 33.5°C. How much heat energy would this require? 100g (33.5°C 22.0°C) = 4810/ ΔΗ = = Suppose a reaction takes place within a calorimeter. All the heat released by the reaction is absorbed by the water solution in the calorimeter and the walls of the calorimeter itself. Thus: -9rxn [9cal + 9soin] ... 8-3 8-Enthalpy Substituting in the heat capacity of the calorimeter (Equation 2) and the specific heat capacity of the solution (Equation 3): -9rxn = [Ccal ATcal] + [Ssoln msoin ATsoin] Finally, insert the numbers used in Example 1 & 2 and solve for the heat released by the reaction: -9rxn = [20 (33.5°C -22.0°C)] + [4.184100g (33.5°C - 22.0°C)] = 5040 J +9rxn=-5040/ The enthalpy change (AH) of this reaction can be solved by scaling heat change (q) to be per mole. qrxn moles of limiting reagent Equation 4: AH is scaled by mole of limiting reagent III. Hess's Law Enthalpy is defined as a state function which means the heat change (at constant pressure) associated with any chemical reaction is always the same, regardless of the path by which the change occurs. An important implication of this is that multiple reactions can be summed together to collectively describe some unknown reaction. Summation of reactions is known Hess's Law. Suppose you would like to calculate the AH associated with the combination reaction between hydrogen gas and carbon to form benzene. (Note that because this is the formation of a compound from its constituent standard state elements, it is the Enthalpy of Formation (AH,) for benzene). 3H₂(g) + 6C(s) → C6H6(l) AH = Unknown To solve for this, three related reactions are known: the combustion of benzene (A), the standard
Transcribed Image Text:Figure 1: Coffee Cup Calorimeter -Greaction=+qsurroundings Equation 1: Calorimetry Assumption Example 1: Heat Capacity Ex. When any substance absorbs heat, it experiences a change in temperature, but every substance has a unique ability to "store" heat. This is called its heat capacity (C), defined as the amount of heat required to raise the temperature of that substance by 1°C. Thus, changing the temperature of the sample from an initial temperature (T) to a final temperature (T) requires heat equal to: q=C.AT Equation 2: Heat transfer as a function of heat capacity and temperature change heat (a), w Example 1: An object with a heat capacity of 20. J/°C has its temperature raised from 22.0°C to 33.5°C. How much heat energy would this require? q=C AT = 20 (33.5°C -22.0°C) = 230 J Reaction Surroundings Insulated Walls Figure 2: Simplified Calorimeter Heat capacity is often scaled to be measured per gram of a substance, a related form known as the specific heat capacity (S), defined as the amount of heat required to raise the temperature of 1 gram of a substance by 1°C. q=S·m AT= 4.184- 4T q=S.m.AT Equation 3: Heat transfer as a function of specific heat capacity, mass, and temperature change gºC Example 2: Specific Heat Capacity Ex. Example 2: A water solution with a specific heat capacity of 4.184 J/g°C and weighing 100. g has its temperature raised from 22.0°C to 33.5°C. How much heat energy would this require? 100g (33.5°C 22.0°C) = 4810/ ΔΗ = = Suppose a reaction takes place within a calorimeter. All the heat released by the reaction is absorbed by the water solution in the calorimeter and the walls of the calorimeter itself. Thus: -9rxn [9cal + 9soin] ... 8-3 8-Enthalpy Substituting in the heat capacity of the calorimeter (Equation 2) and the specific heat capacity of the solution (Equation 3): -9rxn = [Ccal ATcal] + [Ssoln msoin ATsoin] Finally, insert the numbers used in Example 1 & 2 and solve for the heat released by the reaction: -9rxn = [20 (33.5°C -22.0°C)] + [4.184100g (33.5°C - 22.0°C)] = 5040 J +9rxn=-5040/ The enthalpy change (AH) of this reaction can be solved by scaling heat change (q) to be per mole. qrxn moles of limiting reagent Equation 4: AH is scaled by mole of limiting reagent III. Hess's Law Enthalpy is defined as a state function which means the heat change (at constant pressure) associated with any chemical reaction is always the same, regardless of the path by which the change occurs. An important implication of this is that multiple reactions can be summed together to collectively describe some unknown reaction. Summation of reactions is known Hess's Law. Suppose you would like to calculate the AH associated with the combination reaction between hydrogen gas and carbon to form benzene. (Note that because this is the formation of a compound from its constituent standard state elements, it is the Enthalpy of Formation (AH,) for benzene). 3H₂(g) + 6C(s) → C6H6(l) AH = Unknown To solve for this, three related reactions are known: the combustion of benzene (A), the standard
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