Calculate the equilibrium concentrations for all reactants and products if initial concentrations of CH4 is 0.400 M, H2S is 0.800 M, CS2 is 0.400 M, H2 is 0.800 M are mixed and allowed to establish equilibrium.  The Kc for the reaction is 0.0360 and the equilibrium concentration of CH4is 0.556 M.   CH4(g) + 2H2S(g) ↔  CS2(g) + 4H2(g) Group of answer choices [CH4] = 0.556 M, [H2S] = 0.488 M, [CS2] = 0.556 M, [H2] = 1.42 M. [CH4] = 0.556 M, [H2S] = 1.11 M, [CS2] = 0.244 M, [H2] = 0.176 M. [CH4] = 0.556 M, [H2S] = 0.244 M, [CS2] = 0.956 M, [H2] = 1.36 M. [CH4] = 0.556 M, [H2S] = 1.11 M, [CS2

Suppose a 250. mL flask is filled with 1.2 mol of NO and 0.90 mol of NO 2. The following reaction becomes possible: NO3(g) + NO(g) = 2NO2(g) The equilibrium constant K for this reaction is 0.661 at the temperature of the flask. Calculate the equilibrium molarity of NO 3. Round your answer to two decimal places. Ом ☑

The following reactions start at equilibrium.A stress is added and the position of equilibrium shifts. After the indicated stress is added which of the following reactions will proceed towards products?The added stress is the part listed inside the braces {}   NH3(g) + HCl(g)  NH4Cl(s) {and the volume is doubled} CO2(g) + H2(g)  H2O (g) + CO (g) {and volume is reduced} Cl(s) + 2H2(g)  CH4 {and a catalyst is added} N2(g) + 3H2(g)  2NH3(g) {and the volume is reduced}

Calculating equilibrium composition from an equilibrium constant Suppose a 250, mil. flask is filled with 1.5 mol of NO₂, 0.10 mol of CO and 2.0 mol of CO₂. The following reaction becomes possible: NO₂(g) +CO(g) NO(g) + CO₂(g) The equilibrium constant K for this reaction is 0.226 at the temperature of the flask. Calculate the equilibrium molarity of NO₂. Round your answer to two decimal places. OM Explanation CIU Check DEC 9 X 3 6 tv 8

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Suppose a 500. mL flask is filled with 1.1 mol of NO2, 1.7 mol of NO and 0.90 mol of CO2. The following reaction becomes possible: NO, (g)+CO(g) NO(g) +CO,(s) The equilibrium constant K for this reaction is 0.292 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. м B C ol

1. Given the equilibrium constant values N₂ (g) + ¹/2O₂(g) → N₂O (g) N₂O4 (g) → 2 NO₂ (g) ½ N₂(g) + O₂(g) → NO₂ (g) Kel=2.7 x 10-18 K₂=4.6 x 10-³ K3 = 4.1 x 10⁹ Determine the value of K. for the reaction below: 2 N₂O(g) + 30₂ (g) → 2 N₂O4 (g)

O KINETICS AND EQUILIBRIUM Calculating equilibrium composition from an equilibrium constant Suppose a 500. mL flask is filled with 2.0 mol of NO, and 1.3 mol of NO. The following reaction becomes possible: NO, (g) + NO(g) → 2NO, (g) The equilibrium constant K for this reaction is 6.91 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. Explanation Check 02022 McGraw Hill LLC. All Rights Reserved. Terms of Use |P 21

Suppose a 500. mL flask is filled with 0.20 mol of NO, and 1.3 mol of NO. This reaction becomes possible: 2NO,(g) = 2NO(g)+0,(g) Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of O,. You can leave out the M symbol for molarity. NO, NO 0, initial ? change equilibrium