Calculate the value of the equilibrium constant at 25°C for the reaction,2N2O5(g) → 4NO2(g) + O2(g)by using the thermochemical data in the table below.ΔΗkJ/molAGkJ/mols°SubstanceJ/(mol-K)NO2(g)33.251239.9N2O5(g)11118346O2(g)205.2(а) 3.9 х 10-20(b) 1.0 × 10º(c) 1.2 x 100(d) 4.0 x 105(е) 5.1 х 1011

EQUILIBRIUM CONSTANT At equilibrium, the value of the reaction quotient is called the equilibrium…

Calculate the value of the equilibrium constant at 25°C for the reaction, 2N2O5(g) → 4NO2(g) + O2(g) by using the thermochemical data in the table below. ΔΗ kJ/mol s° AG; kJ/mol Substance J/(mol:K) NO2(g) 33.2 51 239.9 N2O5(g) 11 118 346 O2(g) 205.2 (а) 3.9 х 10-20 (b) 1.0 × 10º (c) 1.2 × 10º (d) 4.0 × 10$ (e) 5.1 × 1011

Calculate the value of the equilibrium constant at 25°C for the reaction, 2N2O5(g) → 4NO2(g) + O2(g) by using the thermochemical data in the table below. ΔΗ kJ/mol s° AG; kJ/mol Substance J/(mol:K) NO2(g) 33.2 51 239.9 N2O5(g) 11 118 346 O2(g) 205.2 (а) 3.9 х 10-20 (b) 1.0 × 10º (c) 1.2 × 10º (d) 4.0 × 10$ (e) 5.1 × 1011

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter14: Chemical Equilibrium

Section: Chapter Questions

Problem 15P

See similar textbooks

Similar questions

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)
Calculate the standard Gibbs free-energy change when SO3 forms from SO2 and O2 at 298 K. Why is sulfur trioxide an important substance to study? (Hint: What happens when it combines with water?)
Silver carbonate, Ag2CO3, is a light yellow compound that decomposes when heated to give silver oxide and carbon dioxide: Ag2CO3(s)Ag2O(s)+CO2(g) A researcher measured the partial pressure of carbon dioxide over a sample of silver carbonate at 220C and found that it was 1.37 atm. Calculate the partial pressure of carbon dioxide at 25C. The standard enthalpies of formation of silver carbonate and silver oxide at 25C are 505.9 kJ/mol and 31.05 kJ/mol, respectively. Make any reasonable assumptions in your calculations. State the assumptions that you make, and note why you think they are reasonable.
Given the following information at 25C, calculate G at 25C for the reaction 2A(g)+B(g)3C(g) Substance Hf(kJ/mol) S(J/molK) A(g) 191 244 B(g) 70.8 300 C(g) 197 164 a 956 kJ b 956 kJ c 346 kJ d 346 kJ e 1.03 103 kJ
Explain how the entropy of the universe increases when an aluminum metal can is made from aluminum ore. Thefirst step is to extract the ore, which is primarily a formof A12O3, from the ground. After it is purified by freeingit from oxides of silicon and iron, aluminum oxide ischanged to the metal by an input of electrical energy. 2Al2O3(s)electricalenergy4Al(s)+3O2(g)
What determines Ssurr for a process? To calculate Ssurr at constant pressure and temperature, we use the following equation: Ssurr = H/T. Why does a minus sign appear in the equation, and why is Ssurr inversely proportional to temperature?
When (if ever) are high temperatures unfavorable to a reaction thermodynamically?
The 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?
Consider the reaction of 2 mol H2(g) at 25C and 1 atm with 1 mol O2(g) at the same temperature and pressure to produce liquid water at these conditions. If this reaction is run in a controlled way to generate work, what is the maximum useful work that can be obtained? How much entropy is produced in this case?
Tetrachloromethane (carbon tetrachloride), CCl4, has a normal boiling point of 76.7C and an enthalpy of vaporization, Hvap, of 29.82 kJ/mol. Estimate the entropy of vaporization, Svap. Estimate the free energy of vaporization, Gvap, at 25C.
Enthalpy changes often help predict whether or not a process will be spontaneous. What type of reaction is more likely to be spontaneous: an exothermic or an endothermic one? Provide two examples that support your assertion and one counterexample.
From data in Appendix J, estimate (a) the boiling point of titanium(IV) chloride. (b) the boiling point of carbon disulfide, CS2, which is a liquid at 25 C and 1 bar.