Question

water and Change WILT ADVANCEU TUPILS-Silberberg/Amatels, Be, Equilibrium: The Extent of Chemical N, 0, (g) = 2NO, (g); K 6.1 X 10 at 298 K If [N,O) at the start of the reaction = 2.0 M, select ALL the options that correctly show how to calculate [NO] at equilibrium. Assume that the change in (NO) - +2x. Check all that apply. K = 6.1 X 1024 x= 2.8% of [N, so the simplifying assumption is justified. Assume that x is small, so that [NO,.. = 2.0 M. 6.1 x 10 x 40,012 2.0 If the change in (NO2) = +2x then [N,O.Loom - (2.0 - X). Do you know the answer? Read about this I know it Think so Unsure No idea 27 items left

Answer 1

Ng Oy (9) = 2NO, (g); kc = 6:1 xlo Cat 288k) initial concentration of No Oy; ie [as Qy]. - 20m [NO] at equilibrium = ? solution, No Oy (0) = 210363) Initially; 210 - at equilib 2.0 x 20 Thus; ke = (No)? 6.1X/63 = [206]? (2.0-x) (6.1.X 60-3 (2-x) = you? 0.0122 - 0.0061 n = yu? 402 + 0.00614 - 0.0/22 - 0 on selving above quadratic equation by using formula; x = - b ± √b² - 4ac 2a we get x = 0.0544 ~ 0.1 ie value of x is very small

D Thu Guix 10-3 = (2x)? (9.0-se) Because value of x is very small, thus it can be neglected with respect to 2.0 in denominator we can write 2.0 ke 6.18 10 3 = 4 x ² This first statement is correct [N, 04] initial = 2.0 M value of x (as shown in previous calculation); 0.0544 we have to check x = 2.8% of [N₂0 y initial : x = 2.8% of Now] initial 2 = 2.8 x 2.0. 0.056 10.1 value of a calculated by quadratic equation, - 2 oil Thus, this statement is also correct موا - 200 = 2(0.1) = 0.2 (MO) agm 7 20m this statement is incorrect. Thus,

Kc = 6.1X103 = 2n) 2. (2.000) 6.1 x 10 3 = 4x2 216.1X 1009) . 22 4 ** / 2 (6.1% 103 This x 7 6.18 10 3 xy 2.0 given statement is incorrect 5) If change in (Nex) = +21, then [ns 043egm =(2:09) This statement is correct; as shown in starting of question [N₂Oy] ? 2002 initially 2.0 - at qm; [2.0-x ] +2x le (N2O Den 2.0-s [No = + 2x
Statement 1, 2 and 5 do apply.