Firefly luciferase is the enzyme that allows fireflies to illuminate their abdomens. Because this light generation is an ATP-requiring reaction, firefly luciferase can be used to test for the presence of ATP. In this way. luciferase cantest for the presence of life. The coupled reactions areluciferin + O, - oxyluciferin + lightАMP + PPiАТРGiven that for the coupled reaction, A,G(278K) = -7.70kJ mol1, and for the hydrolysis of ATP to AMP, A,G(278K) = -31.6 kJ mol-1, what is K for the first reaction at 278 K ?Express your answer numerically.• View Available Hint(s)Templates Symbolsundo redo Tesết keyboard shortcuts helpexporent subscript fraction squaré root nth root vector unit vector absoluteValue sciertific notationdejete Close keyboaK = 8.79 • 10-6 The reactionSO2 (9) + 2H,S(g)= 35(s) + 2H20(g)• Part Cis the basis of a suggested method for removal of SO, from power-plant stack gasesThe standard Gibbs energies of formation and the standard enthalpies of fomation at 298 K aregiven below. The values may be helpful when answering questions about tAssume that the partial pressure of sulfur dioxide, Pao, . is equal to the partial pressure of dihydrogen sulfide, Pi,s, and therefore Pso, = Pa,s- If the vapour pressure of water is 25 Torr , calculate the equilibrium partialpressure of SO2 (Pao,) in the system at 298 K.e process.A, HAGSubstance kJ mol (kJ mol)Express the pressure in bars to two significant figures.• View Available Hint(s)H20(9)-228.6-241.8-237.-285.8Templates SymBols undo redo feset keyboard shortcuts helpSO2(9)-300.4-296.9SO: (9)-370.4-395.2Pso, =barH2S(g)-33.01-20.17S(s)Submit

Answered: Image /qna-images/answer/ea98d89a-2a65-4589-8928-40ba1e608334.jpg

The reaction so: (9) + 2H,S(9) - 35(s) + 2H;0(9) • Part C is the basis of a suggested method for removal of SO, from povver-plant stack gases. The standard Gibbs energies of formation and the standard enthalpies of formation at 208 K are given belov. The values may be helpful when answering questions about the process. Assume that the partial pressure of sulfur diaxide. Po,. is equal to the partial pressure of dihydrogen sulfide, Ri,s. and therefore so, = Ps: If the vapour pressure of water is 25 Torr, calculate the equilibrium partial pressure of SO2 (Pso, ) in the system at 298 K. Ar H (kJ mol) (kJ mol) -241.8 AG Express the pressure in bars to two significant figures. Substance > View Available Hint(s) H;0(g) -228.6 H;0() -237.1 -285.8 Undo rego feset keyboard: shortcuts Help SO2(9) 300.4 290.9 SO:(9) -370.4 395.2 Pso, = bar H;S(g) 33.01 20.17 S(s)

The reaction so: (9) + 2H,S(9) - 35(s) + 2H;0(9) • Part C is the basis of a suggested method for removal of SO, from povver-plant stack gases. The standard Gibbs energies of formation and the standard enthalpies of formation at 208 K are given belov. The values may be helpful when answering questions about the process. Assume that the partial pressure of sulfur diaxide. Po,. is equal to the partial pressure of dihydrogen sulfide, Ri,s. and therefore so, = Ps: If the vapour pressure of water is 25 Torr, calculate the equilibrium partial pressure of SO2 (Pso, ) in the system at 298 K. Ar H (kJ mol) (kJ mol) -241.8 AG Express the pressure in bars to two significant figures. Substance > View Available Hint(s) H;0(g) -228.6 H;0() -237.1 -285.8 Undo rego feset keyboard: shortcuts Help SO2(9) 300.4 290.9 SO:(9) -370.4 395.2 Pso, = bar H;S(g) 33.01 20.17 S(s)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

17.) Calculate the change in the Gibbs energy if one mole of a perfect gas is allowed to expand isothermally and reversibly at 60°C from a volume of 3.5 L to 8.0 L. A. -2.29 kJ B. +2.29 kJ C. +9.32 kJ D. -9.32 kJ 18.) Determine the change in the Gibbs energy for the reaction at 298 K. The Enthalpy of formation for CO2 (g) was -393.51 kJ/mol. The entropies for C (s), 02 (g) and CO2 (g) are 5.740, 205.138, 213.74 J/K mol C (s) + 02 (g) → CO2 A. -1246 kJ/mol B. +1245 kJ/mol C. -459.36 kJ/mol D. -394.4 kJ/mol
For the reaction CaO (s) + CO₂(g) CaCO3(s) AH° = =+bT² - cT + d where a, b, c and d are constants. Derive the expressions for AU, AC, and AC,.
Calculate the standard Gibbs energy of the reaction at 298K, from the standard entropies and enthalpies of formation. CO(g) + CH3OH(l) yield CH3COOH(l)
Please don't provide handwritten solution ....
In each case, she fills a reaction vessel with some mixture of the reactants and products at a constant temperature of 72.0 °C and constant total pressure. Then, she measures the reaction enthalpy AH and reaction entropy AS of the first reaction, and the reaction enthalpy AH and reaction free energy AG of the second reaction. The results of her measurements are shown in the table. Complete the table. That is, calculate AG for the first reaction and AS for the second. (Round your answer to zero decimal places.) Then, decide whether, under the conditions the engineer has set up, the reaction is spontaneous, the reverse reaction is spontaneous, or neither forward nor reverse reaction is spontaneous because the system is at equilibrium. P₂010 (s) + 6H₂0 (1) 4H₂ PO(s) 4 2NH₂ (g) → N₂H₂(g) + H₂ (g) 2 4 2 ΔΗ = -439. kJ AS-1179. AG = AH = k AS = kJ Which is spontaneous? this reaction the reverse reaction neither 188. kJ OK- K J K AG 16. kJ Which is spontaneous? this reaction the reverse…
Calculate the equilibrium constant at 37 degree Celsius for the reaction glucose +6O2 +36ADP +36Phosphate => 6CO2 +42H2O +36ATP Where change in G=-423kcal/mole of Glucose
The equilibrium reaction for the aqueous dissociation of acetic acid is shown below. CH3COOH(aq) = CH3COO-(aq) + H+ (aq) Given the data in the table below and your knowledge of the "chemical standard state" (X) and the "biochemical standard state” (Xº'), answer parts a) to e). Gibbs free energy of formation, in units of kJ mol-1, at T = 298.15 K AfGe A¢G° -396.5 -369.3 0 CH3COOH(aq) CH3COO- (aq) H+ (aq) d) For the dissociation of acetic acid at 298.15 K, calculate A,Gº¹. 0
The reduction of iron(III) oxide (Fe₂O,) to pure iron during the first step of steelmaking, 4 - 4 Fe(s) + 30, (e) 4 2 Fe,0,(3) is driven by the high-temperature combustion of coke, a purified form of coal: C(s) + O₂(e) co₂(e) Suppose at the temperature of a blast furnace the Gibbs free energies of formation AG, of CO, and Fe₂O, are-413. kl/mol and -826. kJ/mel, respectively. Calculate the maximum mass of pure iron that can be produced by the combustion of 6200. kg of cake. Round your answer to 2 significant digits. A kg 0.P 00 1.60 9
how can i solve this?
This question deals with the vapourization of isopropyl alcohol: CH;CH(OH)CH3(1) =CH;CH(OH)CH3(g) Part A The boiling point of isopropyl alcohol is 355.5 K and the Compute the standard entropy of vapourization, Ayaps° of isopropyl alcohol. enthalpy of vaporization, AvapHº = 39.76 kJ mol-1.
This question has multiple parts. Work all the parts to get the most points. Use the References to access important values if needed for this question. The Gibbs Free Energy equation is given by the equation: AG=AG + RTln Q - Where: AG = Gibbs Free Energy change AG = Standard Gibbs Free Energy change T= temperature In Q = natural log of the reaction quotient In order to solve for the temperature, T, in two steps you must: Step One Add the same expression to each side of the equation to leave the term that includes the variable by itself on the right-hand side of the expression: (Be sure that the answer field changes from light yellow to dark yellow before releasing your answer.) +AG = +AG + RT In Q Drag and drop your selection from the following list to complete the answer: 1 -AG AG AGⓇ AGⓇ Submit
For the gas-phase reaction 2SO2 + O2 =2SO3, the observed mole fractions for a certain equilibriummixture at 1000 K and 1767 torr are xSO2 = 0.310, xO2 = 0.250, and xSO3 = 0.440. Calculate KP at 1000 K