For the following reaction at 50°C and at 2 atm pressure, 2N_2O_5(g) 2N_2O_4(g)+O_2(g) N_2O_5 is 50\% dissociated. The magnitude of standard free energy change at this temperature is x. x= ______ J mol^ -1 [Nearest integer]. Given: R=8.314 J mol^ -1 K^ -1 , 2=0.30, 3=0.48, 10=2.303, °C+273=K
A. A
Let the initial moles of N_2O_5 be 2. The given reaction is: 2N_2O_5(g) 2N_2O_4(g) + O_2(g) Since N_2O_5 is 50\% dissociated (degree of dissociation = 0.5), the moles at equilibrium are: Moles of N_2O_5 = 2(1 - 0.5) = 1 Moles of N_2O_4 = 2(0.5) = 1 Moles of O_2 = 0.5 Total moles at equilibrium = 1 + 1 + 0.5 = 2.5 Given the total pressure P = 2 atm, the partial pressures of the gases are: P_ N_2O_5 = 1 2.5 2 = 0.8 atm P_ N_2O_4 = 1 2.5 2 = 0.8 atm P_ O_2 = 0.5 2.5 2 = 0.4 atm The equilibrium constant K_p is: K_p = (P_ N_2O_4 )^2 P_ O_2 (P_ N_2O_5 )^2 = (0.8)^2 0.4 (0.8)^2 = 0.4 The standard free energy change G^ is given by: G^ = -RT K_p Using the given values: (0.4) = 2.303 (0.4) = 2.303 ( 4 - 10) (0.4) = 2.303 (2 2 - 1) = 2.303 (2 0.30 - 1) = 2.303 (-0.4) = -0.9212 Now, substituting the values into the G^ equation (T = 50^ C + 273 = 323 K): G^ = -8.314 323 (-0.9212) G^ = 2473.81 J mol^ -1 The magnitude of the standard free energy change to the nearest integer is 2474 J mol^ -1 . Answer: 2474
Related: Chemistry — Chemical Equilibrium · All PYQ Banks