per mole of water. Two trials were conducted using HCl to form NaCl salt. The heat of neutralization per mole of water was found to be -55 kJ/mol with a percent error of -4.51%. Two trials were conducted using HNO3 to form NaNO3 salt by the two students mentioned in the references section. The heats of neutralization per mole of water were found to be -57.6 kJ/mol‚ with a percent error of -4.51%. Although the percent errors weren’t very large‚ they were still off. These could have occurred due to
Premium Thermodynamics Energy Enthalpy
and third reactions. EVALUATION Following are the results for our experiment; ΔH1 = -46.0 kJ mol-1 ± 8.5 kJ mol-1 ΔH2 = -50.2 kJ mol-1 ± 8.5 kJ mol-1 ΔH3 = -98.3 kJ mol-1 ± 8.4 kJ mol-1 ΔH3 = -96.2 kJ mol-1 ± 17 kJ mol-1 (this is the value that has to be found according to Hess’s Law The literature values for ΔH1 and ΔH2 are provided by our teacher. The values are -40 kJ mol-1 and -57 kJ mol-1 respectively‚ making the
Premium Thermodynamics Enthalpy Heat
CALORIMETRY Abstract: During the experiment‚ the group were able to perform the following objectives; to compute the heat capacity of a Styrofoam-cup calorimeter‚ and also to compute the heat of neutralization of 1.0 M hydrochloric acid and 1.0 M sodium hydroxide‚ the heat of dilution of concentrated sulfuric acid‚ and the heat of solution of solid ammonium chloride The sixth experiment was named "Calorimetry" wherein it is the measurement of how much heat is gained or released by a system
Premium Energy Enthalpy Hydrochloric acid
1/14/2013 Principles of General Chemistry‚ 2nd ed. By M. Silberberg Chemistry‚ 8th ed. by W. Whitten‚ R. Davis‚ R.‚ M. L. Peck‚ and G. Stanley. Chemical Thermodynamics: Heat and Thermochemistry Thermodynamics & Thermochemistry Thermodynamics is the study of heat and its transformations to and from other forms of energy Thermochemistry is a branch of thermodynamics that deals with the heat involved with chemical and physical changes Objectives 1. Define thermodynamics & thermochemistry 2. Define
Premium Energy Thermodynamics Entropy
Aim To see which hydro carbons produce the most energy. Hypothesis Different fuels transfer different amounts of energy when they burn because of the different numbers of carbon atoms in fuel molecules. Prediction I think that paraffin will give off the most energy when burned because it has more carbon atoms and carbon chains so it will be harder to break these chains making it harder to burn. Although it is harder to burn because it has more chains and carbon atoms it will release a lot
Premium Heat Hydrocarbon Gasoline
Molar Heat of Combustion Aim: To find the molar heat of combustion for four different alkanols: 1. Methanol 2. Ethanol 3. 1-Propanol 4. 1-Butanol - And to compare the experimental value with the theoretical. Background: The Molar Heat of Combustion of a substance is the heat liberated when 1 mole of the substance undergoes complete combustion with oxygen at standard atmospheric pressure‚ with the final products being carbon dioxide gas and liquid water. (Ref. “Conquering Chemistry‚ Roland
Premium Specific heat capacity Heat Thermodynamics
estimate the ΔG°rxn for the dissolution of zinc iodate at 25°C. (1) 26.4 kJ/mol (2) 22.9 kJ/mol (3) 34.3 kJ/mol (4) 30.9 kJ/mol (5) 36.0 kJ/mol 3. A saturated solution of Ag2SO4 at 25°C contains 0.032 M Ag+ ions. From this information‚ estimate the ΔG°rxn for the dissolution of silver sulfate at 25°C. (1) 17 kJ/mol (2) 20 kJ/mol (3) 27 kJ/mol (4) 32 kJ/mol (5) 35 kJ/mol 4. When 20.0 mL of 0.100 M NaOH is added to 20.0 mL of 0.400 M NaH2PO4
Premium Concentration Chemistry Thermodynamics
of liquid water into 2 moles of steam‚ an additional 82 kJ of energy is required. By using Hess’s Law‚ ΔH + 2(799) + 2(41) + 4(460) = 4(410) + (496) ΔH = -888 kJ mol-1 Alternative: energy absorbed during bond breaking = 4(410) + 2(496) = 2632 kJ energy released during bond forming = 2(799) + 4(460) + 2(41) = 3520 kJ
Premium Temperature Oxygen Hydrogen
Enhtalpy change - Hess’ law Experiment #7 Date performed: March 4‚ 2014 Presented by: Sam Tabah(I.D.#1402433) & Giuliano Amato(I.D.#1328425) General Chemistry 202 Nya-05 (00006) Vanier college Part A. Objective: The objective of this lab was to determine the standard heat formation (∆H°F) of MgO‚ using a calorimeter and determining the enthalpy of two reactions. Applying Hess’ law we were able to determine the standard heat formation of MgO. Introduction
Free Thermodynamics Enthalpy Energy
Qrxn = -(qcal + qsol) qrxn = -(32.0 + 831) = 863 J 5. Calculate the enthalpy of reaction for the formation of 1 mole of AB in solution. 0.60 M x .025 L = .015 Moles A‚ B‚ and therefore AB (1:1:1) Enthalpy of reaction = -0.831 KJ / .015 moles AB = - 58 KJ/mol. Data Tables: Part 1: Determination of the Heat Capacity of the Calorimeter Initial Temperature 50.0 mL H2O—room temperature: 21.6 °C 50.0 mL H2O—heated: 70.0 °C Mixing Data Time (seconds) Temperature (°C) 20 42.0 40 41
Premium Thermodynamics Temperature Enthalpy