75.0 g of PC_5(g) is introduced into an evacuated 3.00 L vessel and allowed to r

Question

75.0 g of PC_5(g) is introduced into an evacuated 3.00 L vessel and allowed to reach equilibrium at 250 degree C. PCl_5(g) PCl_3(g) + Cl_2(g) If K_p = 1.80 for this reaction, what is the total pressure inside the vessel at equilibrium? 2.88 atm 2.27 atm 4.54 atm 7.42 atm 9.69 atm Which of the following rate laws is consistent with the following mechanism? A(g) + B(g) AB(g) fast equilibrium (K_cl) AB(g) + C(g) rightarrow AC(g) + B(g) slow Rate = k[A][B] Rate = kK_cl[A][B][C] Rate = k[AC][B]/[AB][C] Rate = [AB]/[A][B] Rate = K_cl[AC]/[A][C] The data below refer to the following reaction 2NO(g) + Br_2(g) 2NOBr(g) Find the concentration of Br_2 when the system reaches equilibrium. 4.15 M 4.75 M 4.95 M 5.05 M None of the above

Explanation / Answer

The reaction is PCl5(g)ß-> PCl3(g)+Cl2(g)

Molecular weight of PCl5= 208

Moles of PCl5 in 75gm= 75/208=0.36

Let x= drop in moles concentration to reach equilibrium

So at equilibrium

[PCL5]= 0.36-x, [PCL3]= [Cl2] =x

Total moles at equilibrium = 0.36-x+2x= 0.36+x

From gas law equation, PV= nRT

T= 250 deg.c= 250+273.15= 523.15K, R= 0.0821 L.atm/mole.K, n= 0.36+x

P= (0.36+x)*0.0821*523.15/3 =14.32*(0.36+x) atm

Partial pressure = mole fraction* total pressure = (Moles/ total moles)* total pressure

Partial pressures : PCl5= (0.36-x)/ (0.36+x) *(14.32)*(0.36+x)= (0.36-x)*14.32

PCl3= x*14.32 = Cl2

Kp = [PPCl3] [PCl2] / [PCl5] = x2*14.32/(0.36-x) = 1.8

x2/(0.36-x)= 0.125

when solved, x= 0.1585

So the partial pressures : PCl5= 14.32*(0.36-0.1585)=2.88 atm

And PCl3= PCl2= 2.27

Total pressure = 14.32*(0.36+0.1585)=7.42 atm ( D is correct)

2.

3.

The reaction is 2NO +Br2-----à 2NOBr

Let x= drop in concentration of NOBr to reach equilibrium

At equilibrium

[NO] = 2.5-2x, [Br2] =5-x and [NOBr]= 1+2x

So at equilibrium 2.5-2x= 2

0.5 = 2x, x= 0.5/2= 0.25

So Br2 at equilibrium = 5-0.25= 4.75 ( B is correct )

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