Question

At equilibrium, the concentrations in this system were found to be [N,] = [02] = 0.300 M and [NO] = 0.600 M. N,(g) + O2(g) = 2 NO(g) If more NO is added, bringing its concentration to 0.900 M, what will the final concentration of NO be after equilibrium is re-established? [NO]inal = 0.4

Answer 1

Equilibrium constant: - equilibrium comitant ke express the relationship between the amount of I product and reactants present at equilibrem in reversiable chemical reaction at a given temperature. temper fare I kp ke only depent upon the and pressure at equilibrum. | Relation between kp and ke- Tkp z kc (RT) on ] ke a equilibrium constant in pressure (autmy ke = equilibrium constant in mulle) R = 0.08206 I aton / mol.kar I= temperature inck An= change in mole of product- - Reactant). of gasses state. For Chemical Reaction to - a turut bb g c. Can t db for - Tu foyerostar CASA TB5 1 (c), (DJ, CAJACB) known as uncontration of - E, D, A L B. reatant (componen

Initial at equilibrium! [Na] =[02 19:] = 0.3m - [NO] = 0:6 m. After adding No new [No 1 = 0 gm 1. Chemical Reaction! - case-I concen centim 16 N29) + 02.19) = 2 NO 192 0. 3m 0. 3m 0. 6m ke = Thor wring[N] [02] putting all valuessa 0.600.6 ke =- 0.3.8 0.3 Ike=4 Ti ke depent only temperature; L recana ke will be same after addin Nol. case:-2 let x mole converted into Nz & 029 I N2 (g) + O2(g) + 2 No g. --- 0.3 6.3 0.9 0.3+0 0-3+1 0.9-22 at equilibre

20.9-2012 kc =10-3+m)_(0.3+27 2) 4 = (0.9-2x)2 (0.3+2)2 I take both sides square roots: - > 2 = 0.9-2 R 0.3 tr 0.6 +2x = 0.9-27 La 4x = 0.3 x = og = x = 0.3%9/ I final concentratim of (Nogl -0.9 - 2x =0.9-280-3 - 0.9 - 0.3 2 - [Nolf = 6.75m Answer