Question

Consider the following reaction at equilibrium: 2CO2(g) = 200 (g) + O2(g) AH° = -514 kJ Le Châtelier's principle predicts that a decrease in temperature will decrease the value of the equilibrium constant O increase the value of the equilibrium constant decrease the partial pressure of CO increase the partial pressure of CO2 (g) decrease the partial pressure of O2 (g)

Answer 1

- Le-chatelier's Principle → y gf a system at equlibrium is subjected to strass - Cost by changing T,.P etc. Then system will add Just d himself to such a manner, so as to nutroalized the effect of imposed storess. Arm (top les many people hachd An G = f(T;P, & d (4962) = (sama) dT + (lmpe ) + ( se ST /P, & 8 ds=32 dsa de dan cr) -- An sidT+ tar dp + Ø Gz" d& At equlibrium, dConG) = 0 - Amt dI + Arov dpt Gr" d&= 0) This is the thermodynamic condition telier's Prrinciple. Pon the Le-cha

If the temperature of system (T) change at constant Pi 4pH DI = G" SE .: () pa Hure, 2002(g) = 200 (8) + O2(9), OH°= -51467 4H-514 K7 1. =(-)ve -: (34) =(-) ve so with decrease temperature the extant of . reaction (&) increases. so increase the value of the equilibrium constant. The a equilibrium Shifted to right side. The porward neu favour. B - 2 NH3(8) = N2(8) + 3 H2(8), 14°= +92-4 kJ Kp = (PH2) (PN?) [PNG, PH, and Pitta are the - (PNH3)" - partial pressure] Now, removing the N₂ (8). so Pre decreases. To maintain the value equilibrium constant same. De lowering Particu Pressure of H₂. of the