1) Kc = 45 for the equilibrium : 2 NO2(g) <====> N2O4 (g) at a temperature 85 C
a) What is Kc for N2O42 <====> NO2(g) ?
b) What is Kp for 2 NO2(g) <====> N2O4 (g) at 85 C?
R= 0.0821 L-atm/ k-mol
2) The pH of seawater is 8.10 This corresponds to ________ M H+
1) Kc = 45 for the equilibrium : 2 NO2(g) <====> N2O4 (g) at a temperature 85...
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
A flask is charged with 1.500 atm of N2O4(g) and 0.94 atm NO2(g) at 25°C. The equilibrium reaction is given in the equation below. N2O4(g) 2NO2(g) After equilibrium is reached, the partial pressure of NO2 is 0.512 atm. (a) What is the equilibrium partial pressure of N2O4? ______ atm (b) Calculate the value of Kp for the reaction. ______ (c) Is there sufficient information to calculate Kcfor the reaction? -Yes, because the temperature is specified. -No, because the value of...
At a certain temperature Kc = 9.0 for the equilibrium N2O4(g) U 2NO2. What is Kc at the same temperature for NO2(g) U 1/2 N2O4(g) Please explain step by step
A flask is charged with 1.800 atm of N2O4(g) and 1.00 atm of NO2(g) at 25 ∘C , and the following equilibrium is achieved: N2O4(g)⇌2NO2(g) After equilibrium is reached, the partial pressure of NO2 is 0.519 atm . 1A) What is the partial pressure of N2O4 at equilibrium? 1B) Calculate the value of Kp for the reaction. 1C) Calculate the value of Kc for the reaction.
The Kc for the reaction at 298 K is given N2O4(g) ⇋ 2 NO2(g) Kc = 5.9 × 10–3 What is the Kc for the following reaction? ½ N2O4(g) ⇋ NO2(g) Kc = ?
1. At a particular temperature, K = 2.50 for the reaction: SO2 (g) + NO2 (g) ⇄ SO3 (g) + NO (g). If all four gases had initial concentrations of 1.00 M, calculate the equilibrium concentrations of SO2. 2. At a particular temperature, Kp = 0.25 for the reaction: N2O4 (g) ⇄ 2 NO2 (g). A flask containing only N2O4 at an initial pressure of 4.5 atm is allowed to reach equilibrium. a. Calculate the equilibrium partial pressure of N2O4....
The value of Kc for the the following reaction is 0.470 at 471 K. N2O4(g)---->2NO2(g) Part 1) If a reaction vessel at that temperature initially contains 0.0200 M NO2 and 0.0200 M N2O4, what is the concentration of NO2 at equilibrium? _______M? part 2) What is the concentration of N2O4 at equilibrium? _______M?
A flask is charged with 1.500atm of N2O4(g) and 1.00 atm NO2(g) at 25 degree C, and the following equilibrium is achieved: N2O4(g) 2NO2 After equilibrium is reached, the partial pressure of NO2 is 0.519atm. Calculate the value of Kp for the reaction. Calculate Kc for the reaction.
Consider the equilibrium system: N2O4(g) + 2 NO2(g) for which Kp = 0.1134 at 25 °C and A Hº = 58.03 kJ/mol. Assume that 1 mole of N204 and 2 moles of NO2 are introduced into a 5.0 liter container. What will be the equilibrium value of [N204]? O 0.928 M 0.379 M 0.0822 M 0.358 M 0.042 M
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