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18 Nov 2019
At 600K, K_p = 2.40 Times 10^-8 for the reaction F_2(g) Doubleheadarrow 2 F(g) It a system initially contains 0.500 atm of F_2(g) and no F(g), what will be the pressure of F(g) at equilibrium? A. 1.10 Times 10^-4 atm B. 5.61 Times 10^-5 atm C. 4.06 Times 10^-4 atm D. 2.12 Times 10^-6 atm The equilibrium constant for the reaction 2NOCl(g) Doubleheadarrow 2NO(g) + Cl_2(g) is equal to 0.51 at a certain temperature. What is the equilibrium constant at the same temperature for the reaction NO(g) + 1/2 Cl_2 Doubleheadarrow NOCl(g)? A. 1.4 B. 2.0 C. 3.8 D. 0.71 Consider this reaction, carried out at constant temperature and volume. PCl_5(g) PCl_3(g) + Cl_2(g) How can the position of equilibrium for this reaction be shifted to the right? (A) addition of a catalyst (B) removal of Cl_2 (C) addition of an inert gas at constant volume (D) removal of PCl_5
At 600K, K_p = 2.40 Times 10^-8 for the reaction F_2(g) Doubleheadarrow 2 F(g) It a system initially contains 0.500 atm of F_2(g) and no F(g), what will be the pressure of F(g) at equilibrium? A. 1.10 Times 10^-4 atm B. 5.61 Times 10^-5 atm C. 4.06 Times 10^-4 atm D. 2.12 Times 10^-6 atm The equilibrium constant for the reaction 2NOCl(g) Doubleheadarrow 2NO(g) + Cl_2(g) is equal to 0.51 at a certain temperature. What is the equilibrium constant at the same temperature for the reaction NO(g) + 1/2 Cl_2 Doubleheadarrow NOCl(g)? A. 1.4 B. 2.0 C. 3.8 D. 0.71 Consider this reaction, carried out at constant temperature and volume. PCl_5(g) PCl_3(g) + Cl_2(g) How can the position of equilibrium for this reaction be shifted to the right? (A) addition of a catalyst (B) removal of Cl_2 (C) addition of an inert gas at constant volume (D) removal of PCl_5
Elin HesselLv2
17 Jan 2019