Practice ProblemATTEMPT
Calculate the standard free-energy changes for the following reactions at
Want to see the full answer?
Check out a sample textbook solutionChapter 18 Solutions
Chemistry
- The Gibbs free energy change (∆Go) of reaction A is 300 kJ/mol and that of reaction B is -400 kJ/mol. Which reaction is faster (A, B or impossible to determine with the given information)? Please explain answerarrow_forwardWhich of the following is a quantitative measure of disorder? (correct answer not entropy wrong ) A)Free Energy B)enthalpy C)entropy D) Temperature Which of the following processes is exothermic? (correct answer) A)Rolling a ball up a hill B)Boiling water in a beaker to make steam C)Allowing meat to thaw after taking it out of the freezer D) Reacting hydrogen and oxygen gases to make waterarrow_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
- SIMPLE ALGORITHM: Correct significant figures and rounding off (Conventional). 6. Given the following reactions: CaCO3(s) → CaO(s) + CO2(g) ΔH = 178.1 kJ C(s, graphite) + O2(g) → CO2(g) ΔH = -393.5 kJ Determine the enthalpy of the reaction of CaCO3(s) → CaO(s) + C(s, graphite) + O2(g).arrow_forwardHow do I calculate the overall energy change, ΔG°, for the process A to D? The free energy of D is 2, the free energy of C is 3, and the free energy of B is 17. I used the calculation 2-3-17 = -18 but this is the wrong answer.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
- Select 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_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_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_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/ 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_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
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning