Answer the following questions in terms of thermodynamic principles and concepts of kinetic molecular theory. (a) Consider the reaction represented below, which is spontaneous at 298 K. CO2(g) + 2 NH3(g) → CO(NH2)2(s) + H2O(l); ΔH°298 = −134 kJ (i) For the reaction, indicate whether the standard entropy change, ΔS°298, is positive, or Values for ∆G for a reaction give us a powerful tool to predict if a reaction is possible. We calculate in-situ∆G r using this equation: where The the superscript zero (°) indicates standard state: 25°C (298°K),1 atm Using the equilibrium constant for the reaction (3 above) where the three bidentates replace the six monodentates, it has been found that at a temperature of 25 °C (298 K): ΔG ⦵ = -2.303 RT log 10 (K) = -2.303 R T (18.28 - 8.61) = -54 kJ mol-1 Given the reaction system in a closed container at equilibrium and at a temperature of 298 K: ... Q. Consider the following reaction: ... A. has largest E a. In Chapter 4 we have examined problems of finding the adiabatic reaction temperature in a reactor by energy balance, which will be a reasonable approach except that the reaction coordinate, , has to be left as an unknown. In the previous examples in this Chapter, we have seen several examples for solving through the equilibrium constant. The magnitude of the equilibrium constant is 50 for the temperature at which the data was collected. 3. Using the equilibrium constant calculated in 2, calculate the magnitude of the equilibrium constant for the following reactions at the same temperature. i) 2HI(g) H2(g) + I2(g) The equilibrium constant expression from part b is

What is the value of the equilibrium constant, K c under these conditions? (8 points) [N 2] = 2.89 x 10-2 M, [H 2] = 1.08 x 10-3 M, [NH 3] = 7.73 x 10-6 M . 8. The following data is for component A in a simple Aà 2B type of reaction. Answer the following question about component B based upon this data The reaction quotient, Q, is an expression which deals with initial values instead of the equilibrium value that K deals with. We compare Q and K to determine which direction the reaction will proceed to obtain equilibrium. If Q is greater than K, the s ystem will shift to the left. If Q is less than K, the system will shift to the right. What is the value of the equilibrium constant, K c under these conditions? (8 points) [N 2] = 2.89 x 10-2 M, [H 2] = 1.08 x 10-3 M, [NH 3] = 7.73 x 10-6 M . 8. The following data is for component A in a simple Aà 2B type of reaction. Answer the following question about component B based upon this data Determine ΔG and ΔG° for each of the reactions in the previous problem. Use the data in Standard Electrode (Half-Cell) Potentials to determine the equilibrium constant for the following reactions. Assume 298.15 K if no temperature is given.

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Exercise 3 Calculating the Values of K. The following equilibrium concentrations were observed for the Haber process at 127°C: [NH3] = 3.1 X 10-2 mol/L [N2] = 8.5 X 10-1 mol/L [H 2] = 3.1 X 10-3 mol/L. a. Calculate the value of K at 127°C for this reaction. b. Calculate the value of the equilibrium constant at 127°C for the reaction: 59) Which of the following reactions will have the largest equilibrium constant (K) at 298 K? A) CaCO3(s) → CaO(s) + CO2(g) ΔG° =+131.1 kJ B) 2 Hg(g) + O2(g) → 2 HgO(s) ΔG° = -180.8 kJ C) 3 O2(g) → 2 O3(g) ΔG° = +326 kJ D) Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔG° = -28.0 kJ E) It is not possible to determine without more ... The cell potential is zero at equilibrium (E=0), and Q (the reaction quotient) can now be designated as the equilibrium constant K. Solving for lnK, we obtain: [latex]ln\ K = \frac {nFE^o}{RT}[/latex] Example. Calculate the equilibrium constant K, from the following reaction studied at a temperature of 298K: Consider the following examples: (a) The decomposition of HO into H and O 2 22 at 500 K has a very small equilibrium –constant, Kc = 4.1 × 1048 (b)N(g) + O(g) ⇌ 2NO(g),22at 298 K has Kc = 4.8 ×10–31. • If Kc is in the range of 10 –3 to 103, appreciable concentrations of both reactants and products are present. Day 3: Use the information from the titrations to calculate the equilibrium constant, K c, for this reaction. Preparation: I found it best to place the liquids in burettes around the room to give students easy access for creating their flasks for the equilibrium mixture. 3) Let's start with the first reaction and allow it to go to equilibrium. At that point, we have this: K p 1 = (x) (x) We do not know the pressure values yet, but we do know that they are the same because of the 1:1 stoichiometry between SO 2 and SO 3 in the first reaction. Keep in mind that we are acting like the second reaction has not yet ...

Values for ∆G for a reaction give us a powerful tool to predict if a reaction is possible. We calculate in-situ∆G r using this equation: where The the superscript zero (°) indicates standard state: 25°C (298°K),1 atm The reaction below has an equilibrium constant of K p = 2.26 x 10 4 at 298 K. CO (g) + 2H 2(g) ⇌ CH 3 OH (g) What is the value of K p for the reaction below? AND Will the reaction products or reactants be favored at Equilibrium? 1/2CH 3 OH (g) ⇌ 1/2CO (g) + H 2(g) A) K p = -2.26 x 10 4 ; Products are favored. B) K p = 4.4 x 10 - 4 ...Consider the following reaction at 298 K. 2 SO2(g) + O2(g) 2 SO3(g) An equilibrium mixture contains O2(g) and SO3(g) at partial pressures of 0.50 atm and 2.0 atm, respectively. Using data from Appendix 4, determine the equilibrium partial pressure of SO2 in the mixture. In the appendix I have the following: SO2(g) ΔG(f)= -300kJ/mol . O2(g) Suppose the activation energy of a certain reaction is 250 kJ/mol. If the rate constant at T 1 = 300 K is k 1 and the rate constant at T 2 = 320 K is k 2, then the reaction is __ times faster at 320 K than at 300 K. (Hint: Solve for k 2 /k 1.) (a) 3 x 10-29 (b) 0.067 (c) 15.0 (d) 525 (e) 3 x 10-28. 18. Thermodynamic Data at 298 K INORGANIC SUBSTANCES Substance ... Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G ... Which one of the following will have largest ... Which of the following reactions will have the largest equilibrium constant (K) at 298 K? a) Fe2O3(s) + 3 CO(g) ? 2 Fe(s) + 3 CO2(g) ?G° = -28.0 kJ b) It is not possible to determine without more information. c) 3 O2(g) ? 2 O3(g) ?G° = +326 kJ d) CaCO3(s) ? CaO(s) + CO2(g) ?G° =+131.1 kJ e) 2 Hg(g) + O2(g) ? 2 HgO(s) ?G° = -180.8 kJAll of the following apply to the reaction 2 C(s) \longrightarrow⟶ A(g) + 2 B(g) as it carried out in a sealed rigid container at constant temperature EXCEPT the rate of the reaction decreases. the number of molecules of A decreases. the entropy of the system increases. the total pressure increases. The following equation represents the equilibrium reaction for the dissociation of phosgene gas. COCl2(g) ⇀ ↽ CO(g) + Cl2(g) At 100˚C, the value of Kc for this reaction is 2.2 × 10−8 . The initial concentration of COCl2(g) in a closed container at 100˚C is 1.5 mol/L.

Oct 08, 2008 · consider the following reaction: PbCO3 ->PbO+CO2 and calculate the equilibrium pressure of CO2 in the system at 190C Delta G for CO2 is -394.4 in kj/mol, Delta H kj/mol is -393.5, delta S j/mol*k is 213.6 Delta G for PbCO3 is -625.5, Delta H is -699.1, Delta S is 131.0, Delta G for PbO is -187.9, Delta H is -217.3, Delta S is 68.70 i'm not sure what i'm doing wrong, i've done the equation ... Jan 09, 2015 · The values of the acid dissociation constants for various acids are usually given to you in an exam, acetic acid's equilibrium constant being #1.8*10^(-5)#; however, if the value is not given to you, you can always use the equilibrium concentrations described in the above equation to solve for #K_a#. If standard enthalpies of formation at 298 K for NH3(g) and H2O(l) are –46.00 kJ/mol and –286.0 kJ/mol respectively, calculate the enthalpy change of the reaction. asked by shana on January 27, 2015 Chemistry Question 9 Unsaved What is the rate law for the following reaction, if the order of the reaction is m, an unknown? , for the reaction at 298 K. Include units with your answer. D. Is the reaction spontaneous under standard conditions at 298 K? Justify your answer. E. Calculate the value of the equilibrium constant,

ANS: c) the equilibrium constant of each acid PAGE: 14.2 22. Which of the following is not true for a solution at 25°C that has a hydroxide concentration of 2.5 × 10–6 M? a) Kw = 1 × 10–14 b) The solution is acidic. c) The solution is basic. d) The [H] is 4 × 10–9 M. e) The Kw is independent of what the solution contains. (a) Calculate the value of the equilibrium constant for the reaction at 298 K. 1.31 x 1043 (b) Calculate the standard entropy change, ∆S°, for the reaction at 298 K. -270 J/K (c) If ClF 3 were produced as a liquid rather than as a gas, how would the sign and magnitude of ∆S for the

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