706 CHAPTER 15 Chemical Equilibrium [15.87] At 700 K, the equilibrium constant f

Question

706 CHAPTER 15 Chemical Equilibrium [15.87] At 700 K, the equilibrium constant for the reaction is K, = 0.76. A flask is charged with 2.00 atm of CC14, which then reaches equilibrium at 700 K. (a) What fraction of the CCl4 is converted into C and Cl,? (b) What are the partial pressures of CCly and Cl2 at equilibrium? [15.88] The reaction PCI,(g)+ Ch(g)PCls(s) has Kp 0.0870 at 300 °C. A flask is charged with 0.50 atm PCl3, 0.50 atm Cl, and 0.20 atm PCl, at this temperature. (a) Use the re- action quotient to determine the direction the reaction must proceed to reach equilibrium. (b) Calculate the equilibrium partial pressures of the gases. (c) What effect will increasing the volume of the system have on the mole fraction of Cl2 in the equilibrium mixture? (d) The reaction is exothermic. What effect will increasing the temperature of the system have on the mole fraction of Cl2 in the equilibrium mixture? 15 .9 [15.89] An equilibrium mixture of H2, 12, and HI at 458 contains 0.112 mol H2, 0.112 mol 12, and 0.775 mol HI in a 5.00-L vessel. What are the equilibrium partial pressures when equilibrium is reestablished following the addition of 0.200 mol of HI? Consider the hypothetical reaction A(g) + 2B(g) [15.90] 2 C(g), for which Kc = 0.25 at a certain temperature. A 1.00-L reaction vessel is loaded with 1.00 mol of compound C, which is allowed to reach equilibrium. Let the variable x represent the number of mol/L of compound A present at equilibrium. (a) In terms of x, what are the equilibrium concentrations of compounds Band C? (b) What limits must be placed on the value of x so that all concentrations are positive? (c) By putting the 15

Explanation / Answer

PCl3 + Cl2 = PCl5

Kp = PCl5 /(PCl3*Cl2)

Kp =0.087

PCl3 = 0.5

Cl2 = 0.50

PCl5 = 0.2

FOR reaction determination

Q =PCl5 /(PCl3*Cl2)

Q =(0.2)/(0.5*0.5) =0.8

since

Q >> K; expect forward reaction

b)

in equilibrium, due to stoichiometric factors (+ for products, - for reactants)

PCl3 = 0.5 -x

Cl2 = 0.50 - x

PCl5 = 0.2 +x

substitute in Kp expressio

0.087 = (0.2-x)/(0.5 -x)^2

0.5^2 - x + x^2 = (0.2/0.087) - (1/0.087)x

0.25 - x + x^2 = 2.2988 - 11.49x

x^2 +10.49x -2.0488 = 0

x = 0.1918

PCl3 = 0.5 -x = 0.5 - 0.1918 = 0.3082

Cl2 = 0.50 - x = 0.5 - 0.1918 = 0.3082

PCl5 = 0.2 +x = 0.2+0.3082= 0.5082

c)

if we increase V; then; this favours P decrease, that is, it favours the least mol formation.

Gas foramtion in products = 1

Gas mol in reactants = 1+1 = 2

then this favours the products,

the concnetration/partial pressure of Cl2 must DECREASE

d)

IF exothermic, and increasing T; this must shift towards REACTANTS

since Cl2 is a reactant, then Cl2 will increase

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