Equilibrium pressures question?
If 5.00⋅10−1 atm of HI gas, 0.00100 atm of H2 gas, and 0.00500 atm of I2 gas are present in a 5.00 L flask, calculate the equilibrium pressures of all 3 gases once equilibrium is established ?
If
2 Answers
PH2=0.0402 atm
PI2,eq=0.0442 atm
PHI,eq=0.4216 atm
Well, from your notes, you seem to have
H2(g) + I2(g) ⇌ 2HI(g)
I 0.00100 0.00500 0.500
C +x +x −2x
E 0.00100+x 0.00500+x 0.500−2x
Here, we have written that the reaction goes in reverse. Now that you have
Kp=100=P2HIPH2PI2
=(0.500−2x)2(0.00100+x)(0.00500+x)
Not much we can do here other than solve it in full;
100=(0.500−2x)25×10−6+0.00600x+x2
Multiply through by the denominator.
5×10−4+0.600x+100x2=(0.500−2x)2
Expand the right-hand side.
5×10−4+0.600x+100x2=0.250−2⋅0.500⋅2x+4x2
Get this into standard quadratic form:
−0.2495+0.600x+100x2=−2x+4x2
96x2+2.600x−0.2495=0
Now,
a=96
b=2.600
c=−0.2495
so that
x=−b±√b2−4ac2a
=−(2.600)±√2.6002−4(96)(−0.2495)2(96)
This, as-given, has:
x=0.0392,−0.0662
When we plug these in, however, only
PH2=0.00100+(0.0392)=0.0402 atm
PI2,eq=0.00500+(0.0392)=0.0442 atm
PHI,eq=0.500−2(0.0392)=0.4216 atm
use an ICE table, then find pressures with
Explanation:
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