Page 255 - Theory and Problems of BEGINNING CHEMISTRY

244 RATES AND EQUILIBRIUM [CHAP. 16

−→
16.33. For the reaction N 2 O 4 ←− 2NO 2 calculate the value of the equilibrium constant if initially 1.00 mol of NO 2 and
1.00 mol of N 2 O 4 are placed in a 1.00-L vessel and at equilibrium 0.75 mol of N 2 O 4 is left in the vessel.
Ans. −→
N 2 O 4 ←− 2NO 2
Initial 1.00 1.00
Change −0.25 +0.50
Equilibrium 0.75 1.50

[NO 2 ] 2 (1.50) 2
K = = = 3.0
[N 2 O 4 ] 0.75

16.34. In a 3.00-L vessel at 488 C, H 2 and I 2 (g) react to form HI with K = 50. If 0.150 mol of each reactant is used, how
◦
many moles of I 2 remain at equilibrium?
Ans. The initial concentration of each reagent is 0.150 mol/3.00 L = 0.0500 mol/L. Let the equilibrium concen-
tration of HI be 2x for convenience.
−→
H 2 + I 2 ←− 2HI
Initial 0.0500 0.0500 0
Change −x −x +2x
Equilibrium 0.0500 − x 0.0500 − x 2x

[HI] 2 (2x) 2
K = = = 50
[H 2 ][I 2 ] (0.0500 − x) 2
Taking the square root of each side of the last equation yields
2x
= 7.1
0.0500 − x
2x = 7.1(0.0500) − 7.1x
9.1x = 7.1(0.0500)
x = 0.039
[I 2 ] = 0.0500 − 0.039 = 0.011 mol/L

In 3.00 L, there is then 0.033 mol I 2 .

16.35. Calculate the number of moles of Cl 2 produced at equilibrium in a 20.0-L vessel when 2.00 mol of PCl 5 is heated to
◦
250 C. K = 0.041 mol/L.
−→
PCl 5 ←− PCl 3 + Cl 2
[PCl 3 ][Cl 2 ] x 2
Ans. K = 0.041 = =
[PCl 5 ] 0.100 − x
2
x = 0.0041 − 0.041x
2
x + 0.041x − 0.0041 = 0

2
−0.041 + (0.041) + 4(0.0041)
x = = 0.047
2
moles Cl 2 = (0.047 mol/L)(20.0L) = 0.94 mol
16.36. Calculate the equilibrium CO 2 concentration if 0.200 mol of CO and 0.200 mol of H 2 O are placed in a 1.00-L vessel
and allowed to come to equilibrium at 975 C.
◦
−→
CO + H 2 O(g) ←− CO 2 + H 2 K = 0.64