314 jmolk solve for the activation energy ea ea 6797

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Unformatted text preview: 8.314 J/molK) Solve for the Activation Energy (Ea): Ea = 6797 J/mol = 6.80 kJ/mol 6. The following equilibrium reactions have the following Kc values at 1000 K: N2O4(g) ⇄ 2NO2(g) N2(g) + 2O2(g) ⇄ 2NO2(g) Kc’ = 1.5 x 106 Kc’’ = 1.4 x 10-10 Determine the values of Kc for the following reactions at 1000 K: (10 pts) a) NO2(g) ⇄ ½ N2O4(g) b) N2(g) + 2O2(g) ⇄ N2O4(g) 7. Calculate the equilibrium partial pressures of the reactants and products given the initial partial pressures of PCl5 = 0.560 atm and PCl3 = 0.500 atm and the equilibrium partial pressure of Cl2 = 0.536 atm. (10 pts) PCl5(g) ⇄ PCl3(g) + Cl2(g) ; Kp = 23.6 I 0.560 0.500 0 C -x +x +x___ E 0.560-x 0.500+x 0.536 PPCl5 = 0.560 – 0.536 = 0.024 atm PCl2 = 0.536 atm given PPCl3 = 0.500 + 0.536 = 1.036 atm Chemistry 102 Exam 1 Name ___Mr. Perfect___________________________________________ Date _...
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This document was uploaded on 03/19/2014 for the course CHEMISTRY 102 at Los Angeles Harbor College.

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