Question

At 125 degrees C, the equilibrium constant K_p for the following reaction is 0.25.

2 NaHCO₃ (s) $\leftrightharpoons$ Na₂CO₃(s) + CO₂(g) + H₂O(g)

a.) Calculate K_c for the reaction at 125 degrees C

b.) At the beginning of the reaction given in part (A) only 10.0g of NaHCO₃ was there in an evacuated 1.00L flask. What is the total pressure in the flask, after the equilibrium is established 125 degrees C?

Answer 1

Ans The equillibiwn equation is written as. Nahco(s) = Nagco,CS) + COgGg) + H30 Cg) ke for the reaction is 0.25 Relation between kpd ke is kp = kc (RT) Bng ( Ang = No. of moles of gasesous product - No. of moles of ko= kp - (RT) Ang Ang=2 gaseous reactant Kc = 0.25 (0.0891 Lałm ki mort x398k)? &T=125c =)25+2+36 = 398] Kc = 0.25 0.25 1067.70 132.6758) Ka = 2.34 X10-9 0 2 Nahco es) - Has Coz(s) + CO₂ g) + H₂O (g) Initial log o mass - - - Grittal . 10 - 0.1190 mol 84 moles x n (n = 20). Total moles At cowilli moley at equill. Mole fonction - - e de Mole fraction at equil. (Note: Solid substances have no contribution in equillo)

Partial fuegs wie at equillo an (Here, Pr = Total pressure at eqwillibaium. ) Kp = (2) (DPP 0.25 = 5 (P2) (P, 22- 1.00 1 Pp = 1.00 atm. 1 Total pressure at equillibrium is 1 atm.