Interpretation:
An equation as to be written for the given coupled process and its equilibrium constant should be determined.
Concept Information:
Spontaneous process: A process which is initiated by itself, without the help of external energy source is called spontaneous process. All spontaneous process is associated with the decrease in free energy in the system.
First law of
Second law of thermodynamics: According to second law of thermodynamics, the entropy of the universe is the sum of the entropy of the system and the surroundings. For all spontaneous process the entropy of the universe increases.
Irreversible process: Irreversible process is the process in which the system cannot go back to its initial state. In an irreversible process there is an increase in entropy of the system.
Equilibrium process: When the concentration of the reactants and products remains constant with time the process is said to be at equilibrium.
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CHEMISTRY-ALEK 360 ACCES 1 SEMESTER ONL
- The Ostwald process for the commercial production of nitric acid involves three steps: 4NH3(g)+5O2(g)825CPt4NO(g)+6H2O(g)2NO(g)+O2(g)2NO2(g)3NO2(g)+H2O(l)2HNO3(l)+NO(g) a. Calculate H, S,G and K (at 298 K) for each of the three steps in the Ostwald process (see Appendix 4). b. Calculate the equilibrium constant for the first step at 825C, assuming H and S do not depend on temperature. c. Is there a thermodynamic reason for the high temperature in the first step, assuming standard conditions?arrow_forwarda Calculate K1, at 25C for sulfurous acid: H2SO3(aq)H+(aq)+HSO3(aq) b Which thermodynamic factor is the most significant in accounting for the fact that sulfurous acid is a weak acid? Why?arrow_forwardWhat information can be determined from G for a reaction? Does one get the same information from G, the standard free energy change? G allows determination of the equilibrium constant K for a reaction. How? How can one estimate the value of K at temperatures other than 25C for a reaction? How can one estimate the temperature where K = 1 for a reaction? Do all reactions have a specific temperature where K = 1?arrow_forward
- Adenosine triphosphate, ATP, is used as a free-energy source by biological cells. (See the essay on page 624.) ATP hydrolyzes in the presence of enzymes to give ADP: ATP(aq)+H2O(l)ADP(aq)+H2PO4(aq);G=30.5kJ/molat25C Consider a hypothetical biochemical reaction of molecule A to give molecule B: A(aq)B(aq);G=+15.0kJ/molat25C Calculate the ratio [B]/[A] at 25C at equilibrium. Now consider this reaction coupled to the reaction for the hydrolysis of ATP: A(aq)+ATP(aq)+H2O(l)B(aq)+ADP(aq)+H2PO4(aq) If a cell maintains a high ratio of ATP to ADP and H2PO4 by continuously making ATP, the conversion of A to B can be made highly spontaneous. A characteristic value of this ratio is [ATP][ADP][H2PO4]=500 Calculate the ratio [B][A] in this case and compare it with the uncoupled reaction. Compared with the uncoupled reaction, how much larger is this ratio when coupled to the hydrolysis of ATP?arrow_forwardThe decomposition of diamond to graphite [C(diamond) C(graphite)] is thermodynamically favored, but occurs slowly at room temperature. a. Use fG values from Appendix L to calculate rG and Keq for the reaction under standard conditions and 298.15 K. b. Use fH and S values from Appendix L to estimate rG and Keq for the reaction at 1000 K. Assume that enthalpy and entropy values are valid at these temperatures. Does heating shift the equilibrium toward the formation of diamond or graphite? c. Why is the formation of diamond favored at high pressures? d. The phase diagram shows that diamond is thermodynamically favored over graphite at 20,000 atmospheres pressure (about 2 GPa) at room temperature. Why is this conversion actually done at much higher temperatures and pressures?arrow_forwardA crucial reaction for the production of synthetic fuels is the production of H2 by the reaction of coal with steam. The chemical reaction is C(s) + H2O(g) CO(g) + H2(g) (a) Calculate rG for this reaction at 25 C, assuming C(s) is graphite. (b) Calculate Kp for the reaction at 25 C. (c) Is the reaction predicted to be product-favored at equilibrium at 25 C? If not, at what temperature will it become so?arrow_forward
- What is the sign of the standard Gibbs free-energy change at low temperatures and at high temperatures for the explosive decomposition of TNT? Use your knowledge of TNT and the chemical equation, particularly the phases, to answer this question. (Thermodynamic data for TNT are not in Appendix G.) 2C7H5N3O6(s) 3N2(g) + 5H2O() + 7C(s) + 7CO(g)arrow_forwarda Calculate K1, at 25C for phosphoric acid: H3PO4(aq)H+(aq)+H2PO4(aq) b Which thermodynamic factor is the most significant in accounting for the fact that phosphoric acid is a weak acid? Why ?arrow_forwardThe equilibrium constant for a certain reaction increases by a factor of 6.67 when the temperature is increased from 300.0 K to 350.0 K. Calculate the standard change in enthalpy (H) for this reaction (assuming H is temperature-independent).arrow_forward
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