Calculate the Entropy of the reaction equilibrium between 2NO2 - N2O4 given the equation: Delta G...
Question text Calculate the equilibrium constants, KpKp and KcKc for the equilibrium reaction N2O4(g)⇄2NO2(g)N2O4(g)⇄2NO2(g) at 298 K. N2O4(g)N2O4(g) NO2(g)NO2(g) S0S0 (J/K/mol) 304.29 240.06 ΔfH0ΔfH0 (kJ/mol) 9.16 33.18 Select one or more: A. Kp=9.23Kp=9.23 , Kc=12.3Kc=12.3 B. Kp=0.563Kp=0.563 , Kc=0.33Kc=0.33 C. Kp=0.144Kp=0.144 , Kc=0.0058Kc=0.0058 D. Kp=0.355Kp=0.355 , Kc=1.23
Decomposition of nitrogen dioxide dimer N2O4 is described by the reaction: N2O4(g) = 2NO2(g) Concentration of N2O4 became 2 times less after 2,5⋅103 s. You have to calculate: a) the value of rate constant k of the reaction; b) the value of equilibrium constant Kp. You are given the value of standard Gibb’s energy of formation Goform: substance Goform, kJ/mol NO2(g) 51.6 N2O4(g) 98.4
1. For the equilibrium reaction: 2NO2 - N2O4 a) calculate the literature values of ΔG°, ΔH° and ΔS° for the equilibrium reaction (ΔS° NO2 - 240.4 Jk/mol, N2O4- 304.3 Jk/mol) (ΔG° NO2- 51.839 KJ/mol, N2O4-98.286 KJ/mol) (ΔH° NO2 -33.85 KJ/mol, N2O4- 9.660 KJ/mol) b) From your understanding of the principals of FTIR spectroscopy, why is it that you do not see absorbance peaks for N2 gas but you do see them for NO2 and N2O4?
Calculate the standard entropy, Delta S_rxn^degree, of the following reaction at 25.0 degree C using the data in this table. The standard enthalpy of the reaction, Delta H_rxn^degree, is -633.1 kJ middot mol^-1. 3C_2 H_2(t) rightarrow C_6 H_6 (l) Delta S_rxn^degree = Number J middot K^-1 middot mol^-1 Then, calculate the standard Gibbs free energy of the reaction, Delta G_rxn^degree. Delta G_rxn^degree = Number kJ middot mol^-1 Finally, determine which direction the reaction is spontaneous as written at 25.0 degree...
what happens at the molecular level when the following reaction occurs? N2O4(g)>2NO2(g) What change in entropy and enthalpy occur for this reaction? - Both entropy and enthalpy will increase -Entropy will increase, but enthalpy will decrease -Entropy will decrease, but enthalpy will increase -Both entropy and enthalpy will decrease Explain
For a given reaction, Delta H = -19.9 kJ/mol and Delta S = -55.5 J/K-mol. Calculate the temperature in K where Delta G = 0. Assume that Delta H and Delta S do not vary with temperature. Also, what is the equilibrium constant at that temperature?
Consider the reaction: N2(g) + 3H2(g) <---> 2NH3(g) The Gibbs free energy of formation ((delta)Gfo) for ammonia is -16.5 kJ/mol and the reaction is exothermic. Calculate the (delta)Gorxn and the equilibrium constant for the reaction and clearly state whether K increases or decreases with temperature.
Question 5 The spontaneity of a reaction depends both on the enthalpy change, Delta H, and entropy change, Delta S. Reactions that release energy produce more stable products, and the universe tends toward disorder. Thus, an exothermic reaction with a positive entropy change will always be spontaneous. Mathematically, this relationship can be represented as where Delta G is the change in Gibbs free energy and T is the Kelvin temperature. If Delta G is negative, then the reaction is spontaneous....
A reaction is at equilibrium at 298 K. At 310 K, the Gibbs free energy for the reaction is –12.6 kJ/mol. Assuming that both entropy and enthalpy are independent of temperature, what are the values of the entropy and enthalpy for this reaction?
For the following reaction : 2NO2(g) ? N2O4(g) ?H� = -58.04 kJ/mol Try to predict what happens to the system at equilibrium if: a/ The temperature is raised? b/ The pressure of the system is increased? c/ An inert gas is added to the system at constant pressure? d/ An inert gas is added to the system at constant volume? e/ A catalyst is added to the system?