And, because the new equilibrium expression is the product of the individual exp
ID: 550101 • Letter: A
Question
And, because the new equilibrium expression is the product of the individual expressions, the new urm constant is the product of the individual constants. the individual constants. Therefore, for the reaction at 500 K, K,-( 1.6 × 10+)(2.4 × 1047)-3.8 x 10" Using hypothetical equilibria, Table 15.2 summarizes the various way ions can be manipulated and the corresponding changes t expression and equilibrium constant. that must be made to the equilibrium sponding changes to their equilibrium constants. 15.-4 shows how to manipulate chemical equations and make the corre- Worked Example 15.4 The following reactions have the indicated equilibrium constants at 10 Br2(g) + Cl2(g) ± 2BrCI(g) K,=7.2 Determine the value of K, for the following reactions at 100°C: (a) 2NO(g) + Br(r) = ± 2NOBr(g) (b) 4NOBr(g) + 4N0(g) + 2Br2(g) (d) 2NOBr(g) + ci,(g) 2NO(g) + 2BrCI(g) (c) NOBrig) NO(g) + Br2(g) " trategy Begin by writing the equilibrium expressions for the reactions that are given. Then, etermine the relationship of each equation's equilibrium expression to the equilibrium expression of e original equations, and make the corresponding change to the equilibrium constant for each. tup The equilibrium expressions for the reactions that are given are BrCIl INOBri] his equation is the reverse of original equation 1. Its equilibrium expression is the reciprocal or the original equation: NOBrExplanation / Answer
THE VALUE FOR THE PART E IS AS FOLLOWS :- Kc= [NOBr] x [Cl]^0.5 / [NO] [BrCl] , using the cocept that the equilibrium constant is the concentration of products / concentration of reactant each raised to the power equal to stochiometric coefficient according to balanced chemical equation