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A student ran the following reaction in the laboratory at 494 K: PC13(g) =PC13(g) + C12(E)...
A student ran the following reaction in the laboratory at 479 K PC13(E) + Cl2) PC (e) When she introduced PCIy(R) and Cl() into a 1.00 L evacuated container, so that the initial partial pressure of PCI, was 2.39 atm and the initial partial pressure of Cl; was 1.94 atm, she found that the equilibrium partial pressure of Cl, was 0.373 am Calculate the equilibrium constant, Ky she obtained for this reaction.
1. A student ran the following reaction in the laboratory at 304 K: 2NO(g) + Br2(g) 2NOBr(g) When she introduced NO(g) and Br2(g) into a 1.00 L evacuated container, so that the initial partial pressure of NO was 1.22 atm and the initial partial pressure of Br2 was 0.574 atm, she found that the equilibrium partial pressure of Br2 was 0.246 atm. Calculate the equilibrium constant, Kp, she obtained for this reaction. Kp =
A student ran the following reaction in the laboratory at 312 K: 2NO(g) + Br2(g) 2NOBr(g) When he introduced NO(g) and Br2(g) into a 1.00 L evacuated container, so that the initial partial pressure of NO was 1.07 atm and the initial partial pressure of Br2 was 0.448 atm, he found that the equilibrium partial pressure of Br2 was 0.189 atm. Calculate the equilibrium constant, Kp, she obtained for this reaction. Kp =
A student ran the following reaction in the laboratory at 684 K: H2(g) + 12(g) 22HI(g) When she introduced H2(g) and Iz() into a 1.00 L evacuated container, so that the initial partial pressure of H2 was 4.88 atm and the initial partial pressure of I was 3.74 atm, she found that the equilibrium partial pressure of 12 was 0.465 atm. Calculate the equilibrium constant, Kp. she obtained for this reaction. Kp = Submit Answer Retry Entire Group 9 more...
A student ran the following reaction in the laboratory at 677 K: 2HI(g) H_2(g) + I_2(g) When she introduced HI(g) at a pressure of 5.32 atm into a 1.00 L evacuated container, she found the equilibrium partial pressure of HI(g) to be 4.22 atm. Calculate the equilibrium constant, K_p, she obtained for this reaction. K_p =
A student ran the following reaction in the laboratory at 690. K: H2(g) +12(g) 22HI(g) When he introduced H2(g) and 12(g) into a 1.00 L evacuated container, so that the initial partial pressure of H2 was 4.40 atm and the initial partial pressure of I was 3.28 atm, he found that the equilibrium partial pressure of HI was 5.78 atm. Calculate the equilibrium constant, Kp, he obtained for this reaction. Kp = Submit Answer Retry Entire Group 9 more group...
A student ran the following reaction in the laboratory at 672 K 2NH3(g) = N2(g) + 3H2(8) When he introduced NH3(g) at a pressure of 0.722 atm into a 1.00 L evacuated container, he found the equilibrium partial pressure of NH3(g) to be 8.99x103 atm. Calculate the equilibrium constant, Ky, he obtained for this reaction Rp
1. A student ran the following reaction in the laboratory at 632 K: 2HI(g) ->H2(g) + I2(g) When she introduced 0.362 moles of HI(g) into a 1.00 liter container, she found the equilibrium concentration of I2(g) to be 3.55×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc = 2. A student ran the following reaction in the laboratory at 616 K: CO(g) + Cl2(g) -> COCl2(g) When she introduced 0.131 moles of CO(g) and 0.161 moles...
A student ran the following reaction in the laboratory at 541 K: COC12(E) CO(g) + Cl2(g) When she introduced 1.13 moles of COC12(e) into a 1.00 liter container, she found the equilibrium concentration of Cl2(g) to be 3.83x10-2 M. Calculate the equilibrium constant, K, she obtained for this reaction.
A student ran the following reaction in the laboratory at 425 K: PCl5(g) --> PCl3(g) + Cl2(g) When she introduced 4.59 moles of PCl5(g) into a 1.00 liter container, she found the equilibrium concentration of Cl2(g) to be 3.94×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction.