Suppose a 250. mL flask is filled with 1.9 mol of O, and 1.5 mol of NO. The following reaction becomes possible: N, (g) +O,(g) – 2NO(g) The equilibrium constant K for this reaction is 8.51 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. OM ?

Suppose a 500. mL flask is filled with 1.6 mol of Br,, 1.9 mol of OCl, and 1.7 mol of BrCl. The following reaction becomes possible: Br, (g) +OCl, (g) BROC1 (g) +BrC1(g) The equilibrium constant K for this reaction is 0.630 at the temperature of the flask. Calculate the equilibrium molarity of OCl,. Round your answer to two decimal places. | M

Suppose a 500. mL flask is filled with 0.70 mol of Cl, and 0.90 mol of HCl. The following reaction becomes possible: H, (g)+ Cl, (g) = 2HC1 (g) The equilibrium constant K for this reaction is 8.22 at the temperature of the flask. Calculate the equilibrium molarity of Cl,. Round your answer to two decimal places. OM

Suppose a 500. mL flask is filled with 1.9 mol of H, and 1.1 mol of NH,. This reaction becomes possible: N, (g) +3H,(g) =2NH, (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 N,. You can leave out the M symbol for molarity. olo Ar N2 H, NH, initial change equilibrium 미

Suppose a 500. mL flask is filled with 0.90 mol of NO, 1.6 mol of NO and 1.4 mol of CO,. The following reaction becomes possible: NO,(g) + CO (g) = NO (g)+ CO,(g) The equilibrium constant K for this reaction is 3.12 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. IM

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

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 250. mL flask is filled with 1.0 mol of N, and 2.0 mol of NH,. This reaction becomes possible: N2 (g) + 3H,(g) – 2NH, (g) 2NH3 (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 N,. You can leave out the M symbol for molarity. N2 H, NH, initial ? change equilibrium

Suppose a 500. mL flask is filled with 1.7 mol of OCl2, 0.70 mol of BrOCI and 1.1 mol of BrCl. The following reaction becomes possible: Br2(g) +OCl2(g) BrOCI(g) + BrCl (g) The equilibrium constant K for this reaction is 3.84 at the temperature of the flask. Calculate the equilibrium molarity of OCl2. Round your answer to two decimal places. ☐ M ☑

Suppose a  250.mL  flask is filled with  1.9mol  of  Br2 ,  1.4mol  of  OCl2  and  0.30mol  of  BrCl . The following reaction becomes possible: +Br2gOCl2g    +BrOClgBrClg The equilibrium constant  K  for this reaction is  2.78  at the temperature of the flask. Calculate the equilibrium molarity of  BrOCl . Round your answer to two decimal places.