At 25°C, the following reaction has a Ke value of 0.090 H₂O(g) + Cl₂O(g) — What is Kc for the reverse reaction at the sa
Posted: Thu Jun 09, 2022 8:17 am
- At 25 C The Following Reaction Has A Ke Value Of 0 090 H O G Cl O G What Is Kc For The Reverse Reaction At The Sa 1 (38.93 KiB) Viewed 69 times
1. The equilibrium constant for the following reaction is 0.090: H2O(g) + Cl2O(g) 2 HOCK(0) If 3.0 mol of H₂O gas and 3.0 mol of Cl₂O gas are placed in a 2.00 L container and equilibrium is established. a) What is the equilibrium concentration of HOCI gas? Clearly show your steps? [4 marks] b) Calculate the percent reaction [2 marks] NOTE: percent reaction means percent yield. So this is a reflection of how much product you get vs how much you would have gotten if reaction goes to 100% completion.
If the K₂ of a weak acid is 1.6 x 10-8, the K₁ of its conjugate base must be which of the following? (SATP) a. 6.20 d. 6.3 x 10-7 b. 1.0 x 10-14 e. 7.80 C. 6.8 x 10-7 a b C d e
Kw is which of the following? a. the equilibrium constant for water which is always 1.0 x 10- 14 b. Ka x Kb for conjugate acid / base partners @ 25°C / 1bar C. the log[H₂O] @ 25°C d. both a and c e. none of the above a b C d
Which of the following equilibria will shift towards the products (to the right) with increased pressure? (a) C(s) + O2(g) Ž CO2(g) (b) CO(g) + 1/2O2(glmz CO2(g) CO2(g) + CaQ(s) (c) CaCO3(s) Ž (d) CO2(g) + H₂ (e) N2O4(g) CO(g) + H₂O(g) a b C d e (D 2NO2(g)