Practice Problem ATTEMPT
For each of the following, calculate
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CHEMISTRY LOOSELEAF TEXT W/CONNECT >IP
- 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_forwardFor a spontaneous chemical reaction which of the following are possible signs of ΔΔH and ΔΔS? Select one: a. both b and c b. ΔΔH = (+) and ΔΔS = (-) c. ΔΔH = (-) and ΔΔS = (+) d. neither b nor c Clear my choicearrow_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
- Be sure to answer all parts. Consider the combustion of butane gas: C4H10(g) + 13/2 O2(g) → 4CO2(g) + 5H2O(g) (a) Predict the signs of ΔS and ΔH (b) Calculate ΔG at 298 K.arrow_forwardSelect the circumstances under which a reaction would most likely be spontaneous. a. delta H is positive, delta S is positive, and the temperature is low. b. delta H is positive, delta S is negative, and the temperature is low. c. delta H is positive, delta S is negative, and the temperature is high. d. delta H is negative, delta S is negative, and the temperature is high. e. delta H is negative, delta S is positive, and the temperature is high.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_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_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 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_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 picture/ PLEASE ANSWER DIRECTLY WITH THE EXPLANATION.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 picturearrow_forwardUsing ΔG = ΔH – TΔS, explain under what circumstances the following processes would be spontaneous: a. ΔH<0 and ΔS>0 b. ΔH>0 and ΔS<0 c. ΔH<0 and ΔS<0 d. ΔH>0 and ΔS>0arrow_forward
- Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning