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Loose Leaf for Chemistry
- For which of the following is the change in entropy positive? Check all that apply. Explain. A. C 6 H 12 (l) C 6 H 12 (g)B. Dilution of 1.0 mL of 1.0 M NaOH with 90 mL of waterC. . 2 O 2 (g) + 2 SO (g) 2 SO 3 (g)D. grinding a large crystal of NaCl to powderE. Cooling water from room temperature to 15°C.arrow_forward(answer choice) : always, at low pressure, at low temperature, never, at high pressure a. When ΔH is negative and Δ is positive, the reaction will be spontaneous b. When ΔH is positive and ΔS is negative, the reaction will be spontaneous c. When ΔH is negative and ΔS is negative, the reaction will be spontaneous d. When ΔH is positive and ΔS is positive, the reaction will be spontaneousarrow_forwardConsider the combustion of ethane: 2C2H6 + 7O2 --> 4CO2 + 6H2O Substance ΔHf C2H6 -84.7 O2 0 CO2 -393.5 H2O -242.0 What is ΔHcombustion (ΔHreaction) ? Question 2 options: 550.8 kJ -2856.6 kJ 2856.6 kJ -550.8 kJarrow_forward
- SOLVE THE FOLLOWING PROBLEMS. Write down your answers CLEARLY. Thank you! Lesson Topic: THERMODYNAMICS (Spontaneous reactions and Gibbs free energy) 1. Use the given standard entropy values to calculate the standard entropy change to convert hydrogen chloride to chlorine in the given reaction: 4HCl ₍g₎ + O2₍g₎ → 2Cl₂₍g₎ + 2H₂O₍g₎ S°(J/K•mol): HCl ₍g₎ = 186.8 O2₍g₎ = 205 Cl2₍g₎ = 223 H2O₍g₎ = 188.7 2. Predict whether the following reaction leads to an increase or a decrease in entropy. Explain Why. Ag+₍aq₎ + Cl-₍aq₎ → AgCl₍s₎ 3. Calculate ∆G° for the combustion of ethane 2C₂H₆₍g₎ + 7O₂₍g₎ → 4CO₂₍g₎ + 6H₂O₍ₗ₎ With the following ∆G°f: C₂H₆(g = -32.86 kJ/mol CO₂(g) = -394.4 kJ/mol H₂O(l) = -237.2 kJ/mol O₂(g) = 0 4. From the following ∆H and ∆S values, predict whether the reaction would be spontaneous. If not, at what temperature…arrow_forwardH3. Consider a reaction for which the value of ΔG∘ = -9.199 kJ/mol at a 298.15 K. What is the value of Keq for this reaction at this temperature? Please give typed answerarrow_forwardQuick overview of our lesson: Our topic is all about Second Law of Thermodynamics. Gibbs’ free energy, G is defined by G = H – TSwhere H is the enthalpy, T is the temperature (in Kelvins), and S is the entropy. In a chemical reaction,R ↔P (R are reactants and P are products) at aconstant temperature we have ∆G = ∆H – T∆S.If ∆G < 0 the reaction may proceed spontaneously tothe right.If ∆G = 0 the reaction is in equilibrium.If ∆G > 0 the reaction may proceed spontaneously to the left. please do help me with the questions on the picturearrow_forward
- Quick overview of our lesson: Our topic is all about Second Law of Thermodynamics. Gibbs’ free energy, G is defined by G = H – TSwhere H is the enthalpy, T is the temperature (in Kelvins), and S is the entropy. In a chemical reaction,R ↔P (R are reactants and P are products) at aconstant temperature we have ∆G = ∆H – T∆S.If ∆G < 0 the reaction may proceed spontaneously tothe right.If ∆G = 0 the reaction is in equilibrium.If ∆G > 0 the reaction may proceed spontaneously to the left. please do help me with the questions on the picturePLEASE EXPLAIN ALSO PLSarrow_forwardQuick overview of our lesson: Our topic is all about Second Law of Thermodynamics. Gibbs’ free energy, G is defined by G = H – TSwhere H is the enthalpy, T is the temperature (in Kelvins), and S is the entropy. In a chemical reaction,R ↔P (R are reactants and P are products) at aconstant temperature we have ∆G = ∆H – T∆S.If ∆G < 0 the reaction may proceed spontaneously tothe right.If ∆G = 0 the reaction is in equilibrium.If ∆G > 0 the reaction may proceed spontaneously to the left. please do help me with the questions on the picture/ i think it is to be answered in paragraph form .arrow_forwardQuick overview of our lesson: Our topic is all about Second Law of Thermodynamics. Gibbs’ free energy, G is defined by G = H – TSwhere H is the enthalpy, T is the temperature (in Kelvins), and S is the entropy. In a chemical reaction,R ↔P (R are reactants and P are products) at aconstant temperature we have ∆G = ∆H – T∆S.If ∆G < 0 the reaction may proceed spontaneously tothe right.If ∆G = 0 the reaction is in equilibrium.If ∆G > 0 the reaction may proceed spontaneously to the left. please do help me with the questions on the picture/ PLEASE ANSWER DIRECTLY WITH THE EXPLANATION.arrow_forward
- Quick overview of our lesson:Our topic is all about Second Law of Thermodynamics.Gibbs’ free energy, G is defined by G = H – TSwhere H is the enthalpy, T is the temperature (in Kelvins), and S is the entropy. In a chemical reaction,R ↔P (R are reactants and P are products) at aconstant temperature we have ∆G = ∆H – T∆S.If ∆G < 0 the reaction may proceed spontaneously tothe right.If ∆G = 0 the reaction is in equilibrium.If ∆G > 0 the reaction may proceed spontaneously to the left.please do help me with the questions on the picturearrow_forwardANSWER PARTS A, B, C, & D A. Which one of the following processes produces a decrease in the entropy of the system? A) boiling water to form steam B) dissolution of solid KCl in water C) mixing of two gases D) freezing water to form ice E) melting ice to form water B. ΔS is positive for the reaction __________. A) 2H2 (g) + O2 (g) → 2H2O (g) B) 2NO2 (g) → N2O4 (g) C) CO2 (g) → CO2 (s) D) BaF2 (s) → Ba2+ (aq) + 2F- (aq) E) 2Hg (l) + O2 (g) → 2HgO (s) C. For the reaction C2H6 (g) → C2H4 (g) + H2 (g) ΔH° is + 137 kJ/mol and ΔS° is +120 J/K ∙ mol. This reaction is __________. A) spontaneous at all temperatures B) spontaneous only at high temperature C) spontaneous only at low temperature D) nonspontaneous at all temperatures D. Which of these processes lead(s) to an increase in entropy? (treat all gases as ideal). Circle all that apply. A) The pressure of 1 mole of oxygen gas is allowed to double isothermally. B) Carbon dioxide is allowed to expand isothermally to 10 times its…arrow_forwardConsider the reactionHCl(g)+ NH3(g)NH4Cl(s)Using the standard thermodynamic data in the tables linked above, calculate the equilibrium constant for this reaction at 298.15K.ANSWER:arrow_forward
- Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning