4. For the following chemical equilibrium, Kp-4.6 x 10-14 at 25°C, find the value of Kc...
3. Given Kc or Kp for the following reactions, what is the value of Kp or Ke? (a) 12 (g) + Cl2 (a) 22ICI (g): Kc = 2.0 x105 at 25°C (b) N204(g) + 2NO2(0); Kc = 0.90 at 120 °C (c) CaCO3(s) = Cao (s) + CO2 (ox Kp = 1.67 x 102 at 740 °C 4. A container contains an equilibrium mixture of H2 (g), 12(g), and Hl) at 721 K. The concentration of each substance present at...
(1). The equilibrium constant, Kp, for the following reaction is 1.80×10-2 at 698K. 2HI(g) =H2(g) + I2(g) If an equilibrium mixture of the three gases in a 15.5 L container at 698K contains HI at a pressure of 0.399 atm and H2 at a pressure of 0.562 atm, the equilibrium partial pressure of I2 is atm. (2). Consider the following reaction: PCl5(g) =PCl3(g) + Cl2(g) If 1.17×10-3 moles of PCl5, 0.217 moles of PCl3, and 0.351 moles of Cl2 are at...
For the equilibrium: 2 SO3(g) < = > O2(g) + 2 SO2(g) Kp = 0.269 at 625 oC What is Kc at this temperature? Kp = Kc[RT]Δn R = 0.08206 L-atm/mol K
Standard conditions 298.15 K (25 °C) and 1 bar. Thermodynamic Data Conversions Factors I cal 4.184 J 1 bar 0.10 J cm 1 atm 1.01325 bar 1 cm 0.10 J bar 0°C 273.15K Species/ Phase AHP J mol) MnCOs) Mn2 mo K) V° (cm mo) -894.1 -220.75 -167.159 -393.509 167.159 -285.83 85.8 73.6 56.5 213.79 56.5 69.91 31.073 21.0 17.3 24.7892 L mol 17.3 18.068 Cl CO2() Constants HCI R 8.3145 J mol! K-1 H:Op Continuing with the carbonate reaction...
Ke & Equilibrium Pressures and in • The equil. constant, kp, for the following rxn is 10.5 @ 350k 2 CH₂Cl2 cg) - Chucgs + CCl4 cgi If an equil. mixture of the 3 gases in a 13 L container @ 350 K contains CH₂Cl2 e a pressure of 0.558 atm . CHu o a pressure of 0.272 atm, the equil. PP of coly is atm. L Le Châtelier Calculations : [ ] The equil. constant, K, for the following...
number 8 please
QUESTIONS For the equilibrium N204() 2 NO2(8), at 298 K, Kp = 0.15. For this reaction system, it is found that the partial pressure of N204 is 3.2 * 10 atm at equilibrium. What is the partial pressure of NO2 at equilibrium? a. 0.0022 atm b.21 atm c. 4.6 atm d. 0.0048 atm e. 0.069 atm QUESTION 9 Nitrogen trifluoride decomposes to form nitrogen and fluorine gases according to the following equation: 2 NF3(g) = N2(g) +...
(a)
Write the expression for the equilibrium constant (Kc)
(b)
Find Kc, when the equilibrium concentrations for NOBr, NO and Br2
are 0.46 M, 0.1 M and 0.3 M respectively. Report to 3 Significant
figures.
(c)
Find Kp for the abovementioned reaction at 25°C. (R=0.082 L.
atm/mole.K)
(d)
Find Kc, for this reaction: NO(g) + 1/2 Br2 (g) <-> NOBr
(g)
1. Consider this reaction to answer the following questions touteiluna ada se in contain yo. 2NOBr (g) 2NO(g) +...
2) For the equilibrium: 2 SO3(g) <=> 02(g) + 2 SO2(g) Kp = 0.269 at 625 °C What is Ke at this temperature? Kp = K[R R = 0.08206 L-atm/mol K (5pts)
) Given the following reaction at equilibrium, if Kp = 1.10 at 250.0 °C, Kc = ________. PCl5 (g) PCl3 (g) + Cl2 (g) A) 1.10 B) 2.56 × 10 -2 C) 47.2 D) 42.9 E) 3.90 × 10 -6
The equilibrium constant, Kc, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium constant, Kp, is calculated from partial pressures instead of concentrations. These two equilibrium constants are related by the equation Kp=Kc(RT)Δn where R=0.08206 L⋅atm/(K⋅mol), T is the absolute temperature, and Δn is the change in the number of moles of gas (sum moles products - sum moles reactants). For example, consider the reaction N2(g)+3H2(g)⇌2NH3(g) for which Δn=2−(1+3)=−2. A) For the reaction 3A(g)+3B(g)⇌C(g) Kc =...