Extra Equilibrium Practice Problems

Extra Practice Problems for Equilibrium1. Given the following reaction at equilibrium:S 8(s) + 8 O 2(g) ⇔ 8 SO 2(g) + 27.8 kJUsing LeChatelier’s Principle and relating it to the above chemical reaction,fill in the blanks appropriately for each of the following situations. Usethese abbreviations: INC = increase, DEC = decrease, NC = nochange.[S 8 ] [O 2 ] [SO 2 ]a. Remove SO 2 _____ _____ _____b. Decrease volume of container _____ _____ _____c. Add oxygen gas _____ _____ _____d. Increase the temperature _____ _____ _____e. Increase volume of container _____ _____ _____f. Add solid sulfur (S 8 ) _____ _____ _____g. Add a catalyst _____ _____ _____2. Two solutions are mixed. Predict if a precipitate will form if 25.0 mL of0.0020 M silver nitrate (AgNO 3 ) solution is mixed with 25.0 mL of0.0010 M sodium bromate (Na 2 BrO 3 ) solution.K sp AgBrO 3 = 5.30 X 10 -53. Calculate the solubility of iron (III) hydroxide, Fe(OH) 3 if theK sp = 2.79 X 10 -39 .4. Calculate the K sp of magnesium iodide, MgI 2 . The solubility of this compoundis 59.4 g per 100.0 mL water.5. At a particular temperature, the equilibrium constant, K eq , is 3.75 X 10 -6 forthe reaction shown below.SO 2(g) + NO 2(g) ⇔ SO 3(g) + NO (g)If the initial concentrations of SO 2 is 1.00 M and NO 2 is 1.00 M, what arethe equilibrium concentrations of all species?

6. A solution contains a mixture of 0.050 M bromide ion (Cl - ) and0.0055 M iodide ion. If mercury (II) ion (Hg 2+ ) is slowly added to thismixture, which will precipitate first, mercury (II) chlorideor mercury (II) iodide?K sp HgBr 2 = 6.2 X 10 -20 K sp HgI 2 = 9.8 X 10 -297. Calculate the molarity of the following:a. A solution made by dissolving 25.0 grams of nickel (II) chloride, NiCl 2in 500.0 mL of distilled water.b. A solution made by using 50.0 mL of a 0.100 M solution of sodium nitrate(NaNO 3 ) added to 300.0 mL of distilled water.8. Given the following equilibrium reaction, calculate the equilibrium constant(k EQ ) if the equilibrium concentrations are:[SiH 4 ] = 0.0330 M [Cl 2 ] = 0.250M[SiCl 4 ] = 0.0025 M [H 2 ] = 0.0145 MSiH 4(g) + 2 Cl 2(g) ⇔ SiCl 4(g) + 2 H 2(g)

Answers1.[S 8 ] [O 2 ] [SO 2 ]a. Remove SO2 NC DEC DEC then INCb. Decrease volume NC NC NCc. Add O2 gas NC INC then DEC INCd. Increase temp NC INC DECe. Increase volume NC NC NCf. add solid sulfur NC NC NCg. add a catalyst NC NC NC2. Q = 5.0 x 10 -7 , Q