Consider the following reaction: H20(g) + CO(g) = H2 (g) + CO2 (g). Amounts of H20, CO, H2, CO2 are put on a flask such that the system is already at equilibrium. a)  If the CO put originally in the flask has radioactive 14C but the CO2 has not, where will 14C be found after a long, long time? b)  If the CO2 put originally in the flask has radioactive 14C but the CO has not, where will 14C be found after a long, long time?

Say that you mix Fe3+ and SCN- together so that after mixing, the concentration of Fe3+ (aq) equals 0.0015 M , and the concentration of SCN- (aq) equals 5.5 x 10-4 M. You measure the concentration of product, FeSCN2+, after equilibrium forms. It is 7.2 x 10-5 M. Calculate the value of the equilibrium constant, K, for the reaction.  F e 3 + ( a q ) + S C N − ( a q ) ⟷ F e S C N 2 + ( a q ) K = ? Group of answer choices 0.011 9.52 x 10-3 87.3 105

A 3.00-liter flask initially contains 3.00 mol of gas A and 1.50 mol of gas B. Gas A decomposes according to the following reaction:3A  2B + CThe equilibrium concentration of gas C is 0.115 mol/L. Determine the equilibrium concentration of gas A.  a 0.730 M  b 0.115 M  c 0.655 M  d 0.885 M  e 0.345 M

At a particular temperature, K = 3.68 for the reaction SO2(g) + NO2(g) = SO3(g) + NO(g) Calculate the concentrations of all four gases at equilibrium when 0.574 mol of each gas is mixed in a 1.000-L flask. (In other words, the concentration of each gas is 0.574 M.) HOW DO WE GET THERE? We will define the change in the concentrations to reach equilibrium in terms of x. Let x equal the number of moles per liter of SO, and NO, consumed to reach equilibrium. SO2(g) + NO2(g) = SO3(g) + NO(g) Initial: 0.574 0.574 0.574 0.574 Change: -X +x +x -x Equilibrium: 0.574-x 0.574-x 0.574+x 0.574+x Therefore, [SO,]]NO] [SO,]]NO,] (0.574+æ)(0.574+æ) (0.574 – x)(0.574 – x) K = 3.68 = What is the value of x? X =