Slides
... To predict spontaneity, both the enthalpy and the entropy of a process must be known. Entropy (S) of a system is a measure of how spread out or how dispersed the system’s energy is (measured in J/mol·K). ΔS (entropy change for a system) = Sfinal - Sinitial If randomness increases, ΔS is positive. If ...
... To predict spontaneity, both the enthalpy and the entropy of a process must be known. Entropy (S) of a system is a measure of how spread out or how dispersed the system’s energy is (measured in J/mol·K). ΔS (entropy change for a system) = Sfinal - Sinitial If randomness increases, ΔS is positive. If ...
Chapter 1 questions
... An oxide of sulfur contains 40.0% by mass of sulfur. Calculate the empirical formula of the oxide. Q8. Analysis by mass has indicated the following percentage composition by mass of certain compounds. Calculate the empirical formula of each: a) carbon 75.0%, hydrogen 25.0% b) magnesium 60.3%, oxygen ...
... An oxide of sulfur contains 40.0% by mass of sulfur. Calculate the empirical formula of the oxide. Q8. Analysis by mass has indicated the following percentage composition by mass of certain compounds. Calculate the empirical formula of each: a) carbon 75.0%, hydrogen 25.0% b) magnesium 60.3%, oxygen ...
Laboratory Practices from Physical Chemistry
... 13 Determination of acidity constant of weak acid using conductivity measurements ........... 28 15.1 Quinhydrone electrode ..................................................................................................... 33 15.2 Fe3+/Fe2+ electrode ............................................ ...
... 13 Determination of acidity constant of weak acid using conductivity measurements ........... 28 15.1 Quinhydrone electrode ..................................................................................................... 33 15.2 Fe3+/Fe2+ electrode ............................................ ...
Practice Test 3: Answer Key
... A) All collisions of gaseous molecules are perfectly elastic. B) A mole of any gas occupies 22.4 L at STP. *** C) Gas molecules have no attraction for one another. D) The average kinetic energy for molecules is the same for all gases at the same temperature. ...
... A) All collisions of gaseous molecules are perfectly elastic. B) A mole of any gas occupies 22.4 L at STP. *** C) Gas molecules have no attraction for one another. D) The average kinetic energy for molecules is the same for all gases at the same temperature. ...
5.2 Calculations of Enthalpy Changes (SL/HL)
... H in kJ, m in Kg (in aqueous solutions it will be the mass of water which is the same as it’s volume in dm3) C is the specific heat capacity and in an aqueous solution this will be 4.18 kJ kg-1 K-1 Usually we want the enthalpy change per mole H mol-1 of the limiting reagent. In order to measur ...
... H in kJ, m in Kg (in aqueous solutions it will be the mass of water which is the same as it’s volume in dm3) C is the specific heat capacity and in an aqueous solution this will be 4.18 kJ kg-1 K-1 Usually we want the enthalpy change per mole H mol-1 of the limiting reagent. In order to measur ...
6 Thermodynamics
... S[Cu(s)] = 33 J/K·mol S[O2(g)] = 205 J/K·mol S[CuO(s)] = 42 J/K·mol (A) −111.5 J/mol·K (B) −93.5 J/mol·K (C) +111.5 J/mol·K (D) +93.5 J/mol·K 10. In which of the following processes is the standard entropy of reaction, ∆S°, expected to be negative? (A) (NH4)2CO3 (s) → 2 NH3 (g) + H2O (ℓ) + CO2 (g ...
... S[Cu(s)] = 33 J/K·mol S[O2(g)] = 205 J/K·mol S[CuO(s)] = 42 J/K·mol (A) −111.5 J/mol·K (B) −93.5 J/mol·K (C) +111.5 J/mol·K (D) +93.5 J/mol·K 10. In which of the following processes is the standard entropy of reaction, ∆S°, expected to be negative? (A) (NH4)2CO3 (s) → 2 NH3 (g) + H2O (ℓ) + CO2 (g ...
Solution Chemistry and the Hydrosphere
... Use double arrow ( ⇌ ) to indicate incomplete dissociation to form all ions. Acid: CH3CO2H ⇌ CH3CO2−(aq) + H+(aq) Base: NH3 + H2O ⇌ NH4+(aq) + OH−(aq) ...
... Use double arrow ( ⇌ ) to indicate incomplete dissociation to form all ions. Acid: CH3CO2H ⇌ CH3CO2−(aq) + H+(aq) Base: NH3 + H2O ⇌ NH4+(aq) + OH−(aq) ...
Energy Matters - Perth Grammar
... 5 g of marble chips were added to 50 cm3 of 1 mol l−1 hydrochloric acid. State three ways in which the rate of the chemical reaction can be increased. ...
... 5 g of marble chips were added to 50 cm3 of 1 mol l−1 hydrochloric acid. State three ways in which the rate of the chemical reaction can be increased. ...
Sample Exercise 19.1 Identifying Spontaneous Processes
... (a) Equation 19.12 tells us that ΔG° is the sum of the enthalpy term ΔH° and the entropy term –T ΔS°. The temperature dependence of ΔG° comes from the entropy term. We expect ΔS° for this reaction to be negative because the number of molecules of gas is smaller in the products. Because ΔS° is negati ...
... (a) Equation 19.12 tells us that ΔG° is the sum of the enthalpy term ΔH° and the entropy term –T ΔS°. The temperature dependence of ΔG° comes from the entropy term. We expect ΔS° for this reaction to be negative because the number of molecules of gas is smaller in the products. Because ΔS° is negati ...
Synthesis of monoselenanedisulfanediphosphonate by the reaction
... (10 mmol) of ascorbic acid in 75 ml of water; then a solution of potassium dihydrogenmonothiophosphate, KH2PO3S, prepared by dissolution of 3.04 g (20 mmol) of KH2PO3S in 75 ml of water was added into it. The solution (2) was slowly added to the solution (1), and a clear yellowish-green solution of ...
... (10 mmol) of ascorbic acid in 75 ml of water; then a solution of potassium dihydrogenmonothiophosphate, KH2PO3S, prepared by dissolution of 3.04 g (20 mmol) of KH2PO3S in 75 ml of water was added into it. The solution (2) was slowly added to the solution (1), and a clear yellowish-green solution of ...
Slide 1
... (a) Equation 19.12 tells us that ΔG° is the sum of the enthalpy term ΔH° and the entropy term –T ΔS°. The temperature dependence of ΔG° comes from the entropy term. We expect ΔS° for this reaction to be negative because the number of molecules of gas is smaller in the products. Because ΔS° is negati ...
... (a) Equation 19.12 tells us that ΔG° is the sum of the enthalpy term ΔH° and the entropy term –T ΔS°. The temperature dependence of ΔG° comes from the entropy term. We expect ΔS° for this reaction to be negative because the number of molecules of gas is smaller in the products. Because ΔS° is negati ...
19 BROWN Chemical Thermodynamics PPTSExercise
... (a) Equation 19.12 tells us that ΔG° is the sum of the enthalpy term ΔH° and the entropy term –T ΔS°. The temperature dependence of ΔG° comes from the entropy term. We expect ΔS° for this reaction to be negative because the number of molecules of gas is smaller in the products. Because ΔS° is negati ...
... (a) Equation 19.12 tells us that ΔG° is the sum of the enthalpy term ΔH° and the entropy term –T ΔS°. The temperature dependence of ΔG° comes from the entropy term. We expect ΔS° for this reaction to be negative because the number of molecules of gas is smaller in the products. Because ΔS° is negati ...
Buffer Solutions
... Chapter 16, applications of buffers, common ions, titrations, and solubility. ...
... Chapter 16, applications of buffers, common ions, titrations, and solubility. ...
Buffer Solutions
... Chapter 15, applications of buffers, common ions, titrations, and solubility. ...
... Chapter 15, applications of buffers, common ions, titrations, and solubility. ...
chemical change
... Chemical equilibrium is defined as the state in which the concentrations of the reactants and products do not change with time, unless an external force is brought to bear upon the system at equilibrium. We distinguish between physical equilibrium, and chemical equilibrium : Physical equilibrium : w ...
... Chemical equilibrium is defined as the state in which the concentrations of the reactants and products do not change with time, unless an external force is brought to bear upon the system at equilibrium. We distinguish between physical equilibrium, and chemical equilibrium : Physical equilibrium : w ...
IChO 2012
... whereas aluminum fluoride, AlF 3, has a very high melting point (m.p. = 1291 °C). Is the structure and bonding in aluminum fluoride and aluminum bromide likely to be similar to aluminum chloride? B. An Organoaluminum Halide If [(C2H5)2AlCl]2 is treated with NaF, the air-sensitive fluorine analog, [( ...
... whereas aluminum fluoride, AlF 3, has a very high melting point (m.p. = 1291 °C). Is the structure and bonding in aluminum fluoride and aluminum bromide likely to be similar to aluminum chloride? B. An Organoaluminum Halide If [(C2H5)2AlCl]2 is treated with NaF, the air-sensitive fluorine analog, [( ...
1. Bromine exists naturally as a mixture of bromine
... and X-30 (29.974 amu, 3.09% abundance). Calculate the atomic weight of X. A) 28.1 amu B) 54.0 amu C) 29 amu D) 72.7 amu E) 36.2 amu ...
... and X-30 (29.974 amu, 3.09% abundance). Calculate the atomic weight of X. A) 28.1 amu B) 54.0 amu C) 29 amu D) 72.7 amu E) 36.2 amu ...
Stoichiometry, Lab Basics, Reactions
... (where the vapor pressure of water is 21 torr) and a barometric pressure of 757 torr. If 0.555 L of gas is collected, the partial pressure of hydrogen gas is: A) 0.555 x (273+27) B) 0.555 x 757 C) 757 – 27 D) 757 – 21 E) 0.555 x 300 / 273 ____ 17. When a sample of oxygen gas in a closed container of ...
... (where the vapor pressure of water is 21 torr) and a barometric pressure of 757 torr. If 0.555 L of gas is collected, the partial pressure of hydrogen gas is: A) 0.555 x (273+27) B) 0.555 x 757 C) 757 – 27 D) 757 – 21 E) 0.555 x 300 / 273 ____ 17. When a sample of oxygen gas in a closed container of ...
Document
... exothermic reaction - the energy required to break bonds is less than the energy released when bonds form. ie. energy is produced ...
... exothermic reaction - the energy required to break bonds is less than the energy released when bonds form. ie. energy is produced ...
No Slide Title
... General Method For Finding Hrxn Based on the same procedure used in the previous example the following general relationship can be derived Hrxn = [ Hf(products) ] - [ Hf(reactants) ] Notice what this means. If we have a table for formation enthalpies we can find the value for Hrxn for ...
... General Method For Finding Hrxn Based on the same procedure used in the previous example the following general relationship can be derived Hrxn = [ Hf(products) ] - [ Hf(reactants) ] Notice what this means. If we have a table for formation enthalpies we can find the value for Hrxn for ...
avogadro exam 1994 - University of Waterloo
... 9. Which neutral halogen atom has the smallest ionization energy for the first electron removed? ...
... 9. Which neutral halogen atom has the smallest ionization energy for the first electron removed? ...
Spontaniety Worked Examples
... (a) This process is spontaneous. Whenever two objects at different temperatures are brought into contact, heat is transferred from the hotter object to the colder one. (Section 5.1) Thus, heat is transferred from the hot metal to the cooler water. The final temperature, after the metal and water ach ...
... (a) This process is spontaneous. Whenever two objects at different temperatures are brought into contact, heat is transferred from the hotter object to the colder one. (Section 5.1) Thus, heat is transferred from the hot metal to the cooler water. The final temperature, after the metal and water ach ...
Document
... 17. If 23 g of Na are reacted completely with 18 g of water, how many total grams of products are produced? 2 Na + 2 H2O → 2 NaOH + H2 Use Law of conservation of mass. Exactly stoichiometric amounts are reacted together so product mass equals reactant mass. a) 23 g ...
... 17. If 23 g of Na are reacted completely with 18 g of water, how many total grams of products are produced? 2 Na + 2 H2O → 2 NaOH + H2 Use Law of conservation of mass. Exactly stoichiometric amounts are reacted together so product mass equals reactant mass. a) 23 g ...
Experiment 22
... equilibrium system will shift to the left on being heated if the reaction is exothermic (H < 0, Kc goes down). It will shift to the right if the reaction is endothermic (H > 0, Kc goes up). Again, since we can change temperatures very markedly, we can shift equilibria a long, long way. An endother ...
... equilibrium system will shift to the left on being heated if the reaction is exothermic (H < 0, Kc goes down). It will shift to the right if the reaction is endothermic (H > 0, Kc goes up). Again, since we can change temperatures very markedly, we can shift equilibria a long, long way. An endother ...
Review Chapters 4-6 problems Chem 105 Final Sp07
... and multiplying this ratio by 100%. 33. A French scientist named __________ introduced the law of conservation of matter. 34. In the reaction below, how many grams of PF5 can be produced from the reaction of 1.00 g P4 with 1.00 g F2? P4(s) + 10 F2(g) 4 PF5(g) 35. The pH of 1.0 10-5 M HNO3 is ___ ...
... and multiplying this ratio by 100%. 33. A French scientist named __________ introduced the law of conservation of matter. 34. In the reaction below, how many grams of PF5 can be produced from the reaction of 1.00 g P4 with 1.00 g F2? P4(s) + 10 F2(g) 4 PF5(g) 35. The pH of 1.0 10-5 M HNO3 is ___ ...
Thermometric titration
A thermometric titration is one of a number of instrumental titration techniques where endpoints can be located accurately and precisely without a subjective interpretation on the part of the analyst as to their location. Enthalpy change is arguably the most fundamental and universal property of chemical reactions, so the observation of temperature change is a natural choice in monitoring their progress. It is not a new technique, with possibly the first recognizable thermometric titration method reported early in the 20th century (Bell and Cowell, 1913). In spite of its attractive features, and in spite of the considerable research that has been conducted in the field and a large body of applications that have been developed; it has been until now an under-utilized technique in the critical area of industrial process and quality control. Automated potentiometric titration systems have pre-dominated in this area since the 1970s. With the advent of cheap computers able to handle the powerful thermometric titration software, development has now reached the stage where easy to use automated thermometric titration systems can in many cases offer a superior alternative to potentiometric titrimetry.The applications of thermometric titrimetry discussed on this page are by no means exhaustive. The reader is referred to the bibliography for further reading on the subject.