Chemistry Review Manual 2
Chemistry Review Manual 2
Chemistry Review Manual 2
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Substitute the equilibrium partial pressures into the equilibrium constant equation.<br />
This equation simplifies.<br />
K<br />
2 2<br />
pHI<br />
(2 x)<br />
= = = 33<br />
p p (0.100 −x)(0.100 −x)<br />
H2 I2<br />
⎛ 2x<br />
⎞<br />
⎜ ⎟ =33<br />
⎝(0.100 − x)<br />
⎠<br />
2x<br />
= 33 = 5.7<br />
(0.100 − x)<br />
2<br />
4<br />
2x= 5.7 (0.100 −x)<br />
4<br />
0.574<br />
x = = 0.074<br />
7.74<br />
Thus, the final equilibrium partial pressure of HI is<br />
2<br />
p<br />
HI<br />
= 2 x = 2 × 0.074 2 = 0.148 atm<br />
(b)<br />
H 2<br />
+ I 2 <br />
2 HI<br />
Initial 1.00 0.100 0<br />
Change − x − x +2 x<br />
Equilibrium 1.00 − x 0.100 − x 2 x<br />
Substitute the equilibrium partial pressures into the equilibrium constant equation.<br />
K<br />
2 2<br />
pHI<br />
(2 x)<br />
= = = 33<br />
p p (1.00 −x)(0.100 −x)<br />
H2 I2<br />
This equation requires solution of a quadratic equation.<br />
2<br />
4x<br />
− x+<br />
x<br />
2<br />
(0.100 1.10 )<br />
=33<br />
x − x+<br />
x<br />
2 2<br />
4 = 33(0.100 1.10 )<br />
x<br />
− x+ =<br />
2<br />
29 36.<br />
3<br />
3.3 0<br />
2<br />
ax bx c<br />
+ + = 0<br />
Page 48 of 88
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