http://endmemo.com/sconvert/kilopascalbar.php WebIf the reaction mixture contains 0.30 mol of CO, 0.10 mol of H 2, and 0.02 mol of H 2 O, and an unknown amount of CH 4 at equilibrium at 1300 K in a 1L flask, the concentration of CH 4 in the mixture will be - (K c = 3.90 at 1300 K) 1. 5. 85 × 10 2 M 2. 5. 85 × 10-1 M 3. 5. 85 × 10 3 M 4. 5. 85 × 10-2 M

Solved 38 A closed container at 450 K initially contains - Chegg

WebAt 450K, K p=2.0×1010 /bar for the given reaction at equilibrium. 2SO2(g)+O2(g)⇌2SO3(g) What is K c at this temperature? Viewed by: 727 students Solutions ( 1) … WebAt 450K, K p = 2·0 × 10 10 /bar for the given reaction at equilibrium: ... One mole of H 2 O and one mole of CO are taken in a 10 L vessel and heated to 725K. At equilibrium, 40% of water (by mass) reacts with CO according to the equation Calculate the equilibrium constant for the reaction. buy buy baby ft worth tx

At 450K, Kp =2.0×1010/bar for the given reaction at equilibrium.

WebAt 450 K, Kp=2.0×1010/bar for the given reaction at equilibrium. 2SO2(g)+O2(g) ↔2SO3(g) What is Kc at this temperature? Q. At 700 K, the equilibrium constant Kp, for the reaction … WebAt 450K , K p = 2. 0 × 10 10 bar-1 for the given reaction at equilibrium 2 SO 2 (g) + O 2 (g) ⇌ 2 SO 3 (g) What is K c for SO 3 ⇌ SO 2 + 1 2 O 2? Moderate Solution WebDec 2, 2024 · (a) 2SO 2 (g) + O 2 (g) ⇌ 2SO 3 (g) For the given reaction. ∆n g = n p - n r = 2 – 3 = –1. K p = K c (RT) ∆n. K p = 2 × 10 10 bar –1. K c = \(\frac{K_p}{(RT)^{\Delta n}}\) = ( 2 × 10 10 bar –1) (0.0831 L bar K –1 mol 1 × 450 K) = 7.48 × 10 11 L mol –1 (b) For reserve reaction, K c = \(\frac{1}{K_c(forward\,reaction)}\) = \(\frac{1}{748\times10^{11}}\) = 0.134 x … celito redwood city