Question

1) For this reaction: A(aq) ---> B(aq) the change in the standard free enthalpy is 3.39 kJ at 25 oC and 5.34 kJ at 45 oC. Calculate the value of this reaction's equilibrium constant at 75 oC. 2)A container encloses 31.5 g of CO2(g). The partial pressure of CO2 is 2.87 atm and the volume of the container is 31.0 L. What is the average, or root mean square, speed (in m/s) of the CO2 molecules in this container?

Answer 1

31:59 CO PV = nRT P=2.87 atom Tatm = 1:0 1325 x10 Pal V = 31.0L = 0.031 ma n = 31.59 Vs BRT RIT-PV MW 222 3PV 13 x 29080 2.75 Rl/my? x 0.031 MB S- Vn Mw 31.5 x 10°3 kg K Mo O Vars = 926.6 m/s. A caps – o caqs Q = 3.39 kJ at T = 25° 6= 5.34 kg at T=450 Required, kc at 750 19°= - RT enk At 250, Megė R . à 8.314 8295 - 0.255 45c, keq a eles: -5.43 x 10" e 8.314 31988 0.128 9-45) Slope = -2-054 +1.366 Keq en K I he amaus 0.255 -1.366 298 3.36x103 k-1 0:128 - 2.054 386 31845 x10 kt (3.445-3-39x10 , 32.00 ► g = 2.874 x 10 So, at 75c, T = 348 k keq? – 2:054 - In Kanknow wow T3.145 - 2-874) x 103 . celse 3200 > en kope's = -24212 Kirc = 0.054 legger