Suppose a 500.mL flask is filled with 0.90mol of Br2, 0.30mol of OCl2 and 1.5mol of BrCl. The following reaction becomes possible: +Br2gOCl2g +BrOClgBrClg The equilibrium constant K for this reaction is 0.841 at the temperature of the flask. Calculate the equilibrium molarity of OCl2. Round your answer to two decimal places
The balanced chemical equation is

Total volume
An ICE table is made
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| Initial number of moles | 0.90 | 0.30 | 0 | 1.5 |
| Change in number of moles | -x | -x | +x | +x |
| equilibrium number of moles | ![]() |
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| Equilibrium molarity (M) | ![]() |
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The equilibrium constant


Rearrange above equation



Rearrange above equation


This is quadratic equation with solution








Negative value is discarded as concentration cannot be negative

The equilibrium concentration
Suppose a 500.mL flask is filled with 0.90mol of Br2, 0.30mol of OCl2 and 1.5mol of...
Suppose a 500.mL flask is filled with 1.9mol of Br2, 1.5mol of OCl2 and 0.40mol of BrOCl. The following reaction becomes possible: +Br2gOCl2g +BrOClgBrClg The equilibrium constant K for this reaction is 5.17 at the temperature of the flask. Calculate the equilibrium molarity of BrOCl. Round your answer to two decimal places.
Suppose a 500. ml flask is filled with 1.3 mol of Br2, 2.0 mol of OC1, and 1.2 mol of BrCl. The following reaction becomes possible: Br2(g) + OC12(g) = BroCl (g) + BrCl (g) The equilibrium constant K for this reaction is 0.766 at the temperature of the flask. Calculate the equilibrium molarity of OC12. Round your answer to two decimal places. xs ?
A 500 mL flask was filled with 1.5 mol of OCl2, 0.70 mol of BrOCl and 1.9 mol of BrCl. The reaction becomes: Br2+OCl2--> BrOCl+BrCl The equilibrium constant K for the reaction is 0.966. Calculate the equilibrium molarity of Br2.
Suppose a 500. mL flask is filled with 0.90 mol of OC1,, 0.20 mol of BroCl and 1.4 mol of BrCl. The following reaction becomes possible: Br2(g) +OC12(g) =BroCl(g) +BrCl(g) The equilibrium constant K for this reaction is 0.798 at the temperature of the flask. Calculate the equilibrium molarity of Br2. Round your answer to two decimal places. IM xs ?
Suppose a 500 ml flask is filled with 0.10 mol of Bry, 1.6 mol of OCI, and 0.60 mol of BrCl. The following reaction becomes possible: Bry()+OCI (8) -BrOCI () + BrCl() The equilibrium constant K for this reaction is 0.770 at the temperature of the flask Calculate the equilibrium molarity of OCI. Round your answer to two decimal places x x 5 ?
Suppose a 500.mL flask is filled with 0.10mol of NO and 1.2mol of NO2. The following reaction becomes possible: +NO3gNOg 2NO2g The equilibrium constant K for this reaction is 0.541 at the temperature of the flask. Calculate the equilibrium molarity of NO2. Round your answer to two decimal places
Suppose a 500.mL flask is filled with 0.20mol of NO3 and 1.9mol of NO . The following reaction becomes possible: +NO3 (g) + NO (g) ---> 2NO2 (g) The equilibrium constant K for this reaction is 5.74 at the temperature of the flask. Calculate the equilibrium molarity of NO3 . Round your answer to two decimal places.
= Objective Knowledge Check Question 12 Suppose a 250 ml flask is filled with 1.6 mol of Bry, 1.7 mol of Oct, and 1.0 mol of BrCl. The following reaction becomes possible: Br2(8) +OC1,() - BrOCI(g) +BrCl(g) The equilibrium constant K for this reaction is 2.54 at the temperature of the flask. Calculate the equilibrium molarity of Brz. Round your answer to two decimal places. OM X 5 ?
Suppose a 250. ml flask is filled with 1.8 mol of OC1, 0.70 mol of BroCl and 0.50 mol of BrCl. The following reaction becomes possible: Br2(g) + OC12(g) – Brocl(g) +BrCl(g) The equilibrium constant K for this reaction is 1.48 at the temperature of the flask. Calculate the equilibrium molarity of BroCl. Round your answer to two decimal places. x 5 ?
Suppose a 500 ml flask is filled with 1.2 mol of NO, and 0.40 mol of NO the following reaction becomes possible: NO,() +NO(g) - 2NO(g) The equilibrium constant K for this reaction is 7.29 at the temperature of the flask Calculate the equilibrium molarity of NO. Round your answer to two decimal places