13. Under base – catalysed conditions, two molecules of acetone can condense to - form diacetone alcohol. At room temperature (25°C), about 5% of the acetone is converted to diacetone alcohol. Determine the value of AG° for this reaction. OH HO. acetone diacetone alcohol
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- CH4(g) + 2O2(g) ↔ CO2(g) + 2H2O(g) Identify which way the reaction above would shift given the following initial conditions. You must prove it using math. [CH4]0 = 1.00M [O2]0 = 1.00M [CO2]0 = 2.00M [H2O]0 = 2.00MCalculate the ÄE°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide.ÄH°f [CaCO3(s)] = -1206.9 kJ/mol; ÄH°f [CaO(s)] = -635.1 kJ/mol; ÄH°f [CO2(g)] = -393.5 kJ/mol CaCO3(s) CaO(s) + CO2(g) a. -178.3 kJ b. -2235.5 kJ c. 178.3 kJ d. -175.82 kJ e. 175.82 kJFor a reversible bimolecular association reaction A + B <=> AB, what are the units of Kassoc? M sec-1 M-1 sec-1 M
- If Keq for the reaction below at 225°C is 1.87 x 1010, calculate ΔS°rxn. B2 (g) + 2 A (g) → 2 AB (l) ΔH° = –304,200 J Group of answer choices -414.2 J/K -807.5 J/K -13320 J/K -1155 J/K -832.8 J/KWhich conditions would most effectively prevent the following reaction from occurring? NH3 (g) + HCl (g) = NH4Cl (s) \Delta H^0=-177kJΔH0=−177kJ Low temperature and low pressure. Low temperature and high pressure. High temperature and low pressure. High temperature and high pressure. Addition of HCl (g).The value of Keq for the following reaction is 0.75:A (g) + 2B (g) --><-- C (g) + 4D (g)The value of Keq at the same temperature for the reaction below is ________.1/2A (g) + B (g) --><-- 1/2C (g) + 2D (g) A. 1.5 B. 0.87 C. 0.56 D. 0.38 E. 0.75
- Which of these reactions or processes has a negative DS? (D = delta) (A) C2H5OH(l) --> C2H5OH(g) (B) 2NO2(g) --> 2NO(g) + O2(g) (C) NH3(g) --> NH3(aq) (D) H2(g) + F2(g) --> 2HF(g)Calculate ΔS°rxn for the following reaction. The S° for each species is shown below the reaction. C2H2(g) + 2 H2(g) → C2H6(g) S°(J/mol∙K) 200.9 130.7 229.2 Group of answer choices ΔS°rxn = +229.2 J/K ΔS°rxn = +303.3 J/K ΔS°rxn = +560.8 J/K ΔS°rxn = −233.1 J/K ΔS°rxn = −102.4 J/KCan you please help us with number 10 in the picture. we get 1.4 = log A/HA. so we figure this means because it is greater than one, the reaction shifts to the left and therefore A - > H. We are not sure.This is not a graded question as it is a practice question . I am 60 years old and helping my son prepare for the AP exam in a few months. We do questions at the back of the textbook by Zumdahl and Zumdahl