www Combustion of glucose (C H20) is the main source of energy for animal cells: AG (37 °C) =-2872. kJ 6 CO2(g)+ 6 H2O() CH1206(s) +602(g) rxn This energy is generally stored as ATP (adenosine triphosphate) molecules, which can release it when convenient by hydrolysis ("water-assisted in biochemistry: decomposition") into ADP (adenosine diphosphate) molecules and a phosphate anion, often given the symbol P = -35 to 70 kJ хn AG ADP (aq) +P, (aq). ATP(aq) The actual amount of free energy released by the hydrolysis of ATP varies, depending on the exact conditions inside the cell. Suppose under certain conditions hydrolysis of ATP actually releases -44.0 kJ/mol. Calculate the maximum number of ATPS that could be created from ADPS and P, by the combustion of a molecule of glucose. Your answer must be a whole number. ? Check Explanation 2019 McGraw-Hill Education. All Rights Reserved. Terms o Anse End Home Eso F12 F11 F10 Fo F8 F7 F6 F5 F4 F3 FnLock & % X

www Combustion of glucose (C H20) is the main source of energy for animal cells: AG (37 °C) =-2872. kJ 6 CO2(g)+ 6 H2O() CH1206(s) +602(g) rxn This energy is generally stored as ATP (adenosine triphosphate) molecules, which can release it when convenient by hydrolysis ("water-assisted in biochemistry: decomposition") into ADP (adenosine diphosphate) molecules and a phosphate anion, often given the symbol P = -35 to 70 kJ хn AG ADP (aq) +P, (aq). ATP(aq) The actual amount of free energy released by the hydrolysis of ATP varies, depending on the exact conditions inside the cell. Suppose under certain conditions hydrolysis of ATP actually releases -44.0 kJ/mol. Calculate the maximum number of ATPS that could be created from ADPS and P, by the combustion of a molecule of glucose. Your answer must be a whole number. ? Check Explanation 2019 McGraw-Hill Education. All Rights Reserved. Terms o Anse End Home Eso F12 F11 F10 Fo F8 F7 F6 F5 F4 F3 FnLock & % X

Chemistry for Engineering Students

4th Edition

ISBN:9781337398909

Author:Lawrence S. Brown, Tom Holme

Publisher:Lawrence S. Brown, Tom Holme

Chapter10: Entropy And The Second Law Of Thermodynamics

Section: Chapter Questions

Problem 10.86PAE: The enthalpy of vaporization for water is 40.65 kJ mol-1. As a design engineer for a project in a...

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The enthalpy of vaporization for water is 40.65 kJ mol-1. As a design engineer for a project in a desert climate, you are exploring the option of using evaporative cooling. (a) If the air has an average volumetric heat capacity of 0.00130 J cm-3 K-1, what is the minimum mass of water that would need to evaporate in order to cool a 5 m? 5 m room with a 3 m ceiling by 5°F using this method? (b) Is this a spontaneous or nonspontaneous process?
The oxidation of glucose to CO2 and water is a major source of energyin aerobic organisms. It is a reaction favored mainly by a large negativeenthalpy change. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(1) ΔH ° = -2816 kJ/mol ΔS ° = +181 J/mol • K(a) At 37 °C, what is the value for ΔG° ?(b) In the overall reaction of aerobic metabolism of glucose, 32 moles ofATP are produced from ADP for every mole of glucose oxidized. Calculatethe standard state free energy change for the overall reaction whenglucose oxidation is coupled to the formation of ATP at 37 °C.(c) What is the efficiency of the process in terms of the percentage of theavailable free energy change captured in ATP?
Thermodynamics Quantities for Selected Substances at 298.15 K (25⁰C) Substance ∆H⁰f (kJ/mol) ∆G⁰f (kJ/mol) S (J/K-mol) Carbon C2H2(g) 226.7 209.2 200.8 CO2(g) -393.5 -394.4 213.6 Hydrogen H2(g) 0 0 130.58 Oxygen O2(g) 0 0 205.0 H2O(l) -285.83 -237.13 69.91 6. What is the ∆S⁰ of the combustion of acetylene in the presence of excess oxygen yields carbon dioxide and water: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l) in J/K?
Combustion of glucose ( C6H12O6 ) is the main source of energy for animal cells: C6H12O6 ( s ) + 6O2 ( g ) → 6CO2 ( g ) + 6H2O ( l ) =ΔGrxn37°C−2872.kJ This energy is generally stored as ATP (adenosine triphosphate) molecules, which can release it when convenient by hydrolysis ("water-assisted decomposition") into ADP (adenosine diphosphate) molecules and a phosphate anion, often given the symbol Pi in biochemistry: ATP ( aq ) → ADP ( aq ) + Pi ( aq ) =ΔGrxn−35 to 70kJ The actual amount of free energy released by the hydrolysis of ATP varies, depending on the exact conditions inside the cell. Suppose under certain conditions hydrolysis of ATP actually releases −51.3/kJmol . Calculate the maximum number of ATP s that could be created from ADPs and Pi by the combustion of a molecule of glucose. Your answer must be a whole number.
Thermodynamics Quantities for Selected Substances at 298.15 K (25⁰C) Substance ∆H⁰f (kJ/mol) ∆G⁰f (kJ/mol) S (J/K-mol) Carbon C2H2(g) 226.7 209.2 200.8 CO2(g) -393.5 -394.4 213.6 Hydrogen H2(g) 0 0 130.58 Oxygen O2(g) 0 0 205.0 H2O(l) -285.83 -237.13 69.91 What is the ∆S⁰ in the combustion of ethene in the presence of excess oxygen yields carbon dioxide and water: C2H2(g) + 3O2(g) → 2CO2(g) + 2H2O(l) in J/K?
When nitric acid is produced industrially, nitrogen monoxide, NO, is first formed at high temperature. Bakefetr reacts NO on cooling further with oxygen to nitrogen dioxide: 2 NO(g) + O2 ⇌ 2 NO2 (g) Table 1: Thermodynamic data at 25°C. Bond ΔfHom Som Cop,m NO(g) 90.25 210.76 29.34 O2(g) 0.00 205.14 29.36 NO2(g) 33.18 240.06 37.20 1) Calculate (with all relevant intermediate calculations) the standard reaction Gibbs free energy, ΔrG25o, for reaction (1) at 25°C from the data in Table 1 2) Calculate (with all relevant intermediate calculations) the equilibrium constant K25, for reaction (1) at 25°C. 3) Industrially, however, the reaction does not proceed at 25°C but at 500°C. Therefore, calculate (with all relevant intermediate calculations) the standard reaction Gibbs free energy, ΔrG500o, for reaction (1) at 500°C under the assumption that the standard molar heat capacities, Cop, in Table 1 are independent of temperature in the interval [25°C, 500°C]
Calculate the ΔH°f of C6H12O6(s) from the following data: Show the complete and step by step solution ΔH combustion of C6H12O6(s) = -2816kJ/mol ΔH°f of CO2 (g) = -393.5 kJ/mol ΔH°f of H2O (l) = -285.9 kJ/mol Equation: C6H12O6(s) + O2(g) -> CO2(g) + H2O(l)
Calculate the enthalpy change for the reaction C(s) + 2H2 (g)--->CH4(g) from the following data. C(s) + O2 (g) ---> CO2(g). ΔH = – 393.5 kJ/mol H2(g) + 0.5 O2 (g) ---> H2O(l). ΔH = – 285.8 kJ/mol CH4(g) + 2 O2 (g)---> CO2(g) + 2 H2O(l). ΔH = – 890.2 kJ/mol
A pellet of benzoic acid standard was combusted in a bomb calorimeter to determine the experimental calorimeter constant. After 6 mins of monitoring the temperature, the bomb was fired. The following data were obtained. mole of benzoic acid: 0.0049 mol mass of burnt fuse wire: 0.0117 g qwire,surr: 5858 J/g ΔHcombustion of benzoic acid: -3228.0 kJ/mol Δng (combustion of benzoic acid): -0.5 mol R: 8.314 J/mol•K r1: 0ºC/min r2: 0ºC/min Tf: 30.5ºC Ti: 28.4ºC Determine the calorimeter constant (in J/ºC). Report your answer in 2 decimal places.
Using the data below, the ∆S⁰ is -326.3 J/K in the combustion of hydrogen in the presence of excess oxygen yields water: 2H2(g) + O2(g) → 2H2O(l) in J/K. Thermodynamics Quantities for Selected Substances at 298.15 K (25⁰C) Substance ∆H⁰f (kJ/mol) ∆G⁰f (kJ/mol) S (J/K-mol) Hydrogen H2(g) 0 0 130.58 Oxygen O2(g) 0 0 205.0 H2O(l) -285.83 -237.13 69.91 TRUE OR FALSE
Carbon dioxide from the atmosphere "weathers," or dissolves, limestone (CaCO3) by the reactionCaCO3(s) + CO2(g) + H2O(ℓ) → Ca2+(aq) + 2HCO3-(aq)Obtain ΔH° for this reaction using the following data:ΔH°(CaCO3(s)) = -1206.9 kJΔH°(CO2(g)) = -393.5 kJΔH°(H2O(ℓ)) = -285.8 kJΔH°(Ca2+(aq)) = -542.8 kJΔH°( HCO3-(aq)) = -692.0 kJ
18. Calculate the heat of reaction at constant volume for 5.00 g C3H8 ,considering the following chemical reaction: C3H8 (g) + O2(g) --------------------> CO2 (g) + H2O (g) heat of standard formation: O2 = 0.00 kJ / mole, H2O (g) = -245 kJ / mole, CO2 = -318 kJ/mole and C3H8 = -438 kJ / mole C3H8 = 44.09 g/mole Temperature is 33 oC a.412 kJb.-212 kJc.-789 kJd.-169 kJ