312
7 Chemical Equilibrium
which is the same as
4.054α 2 + 0.148α − 0.148 0
Use of the quadratic formula gives
α
−0.148 ±
(0.148) 2 + (4)(4.054)(0.148)
2(4.054)
0.174
We have disregarded the negative root to the quadratic equation, since a negative value
of α is not physically possible if no NO 2 is initially present. The total amount of gas is
(1.000 mol)(1 − α + 2α) 1.174 mol, so the equilibrium total pressure is
P eq
(1.174 mol)
8.3145 J K −1 mol −1
(298.15 K)
0.02446 m 3
1.190 × 10 5 Pa 1.174 atm
It is always the case that when a quadratic equation is solved in a chemistry problem,
only one of the two roots corresponds to a physically possible situation.
We now consider reactions involving pure liquids and solids as well as gases. If the
pressure of the system does not differ very much from P ◦ , then Eq. (6.3-14) gives for
the activity of a pure liquid or solid
a i ≈ 1 (pure liquid or solid near P
◦ )
(7.2-3)
The activities of pure liquids or solids can be omitted from an activity quotient unless
the pressure is very large.
E X A M P L E 7.4
a. Write the equilibrium constant quotient for the reaction CaCO 3 (s) CaO(s) + CO 2 (g).
b. Find the value of the equilibrium constant at 298.15 K and P eq (CO 2 ) at 298.15 K.
Solution
a. Since the activities of the solids are nearly equal to unity,
K
a eq (CaO)a eq (CO 2 )
a eq (CaCO 3 )
≈ a eq (CO 2 ) ≈
P eq (CO 2 )
P ◦
b. From the Gibbs energy changes of formation
∆G ◦ (1)(−603.501 kJ mol −1 ) − 394.389 kJ mol −1
+ (−1)(−1128.79 kJ mol −1 )
130.90 kJ mol −1
7 Chemical Equilibrium
which is the same as
4.054α 2 + 0.148α − 0.148 0
Use of the quadratic formula gives
α
−0.148 ±
(0.148) 2 + (4)(4.054)(0.148)
2(4.054)
0.174
We have disregarded the negative root to the quadratic equation, since a negative value
of α is not physically possible if no NO 2 is initially present. The total amount of gas is
(1.000 mol)(1 − α + 2α) 1.174 mol, so the equilibrium total pressure is
P eq
(1.174 mol)
8.3145 J K −1 mol −1
(298.15 K)
0.02446 m 3
1.190 × 10 5 Pa 1.174 atm
It is always the case that when a quadratic equation is solved in a chemistry problem,
only one of the two roots corresponds to a physically possible situation.
We now consider reactions involving pure liquids and solids as well as gases. If the
pressure of the system does not differ very much from P ◦ , then Eq. (6.3-14) gives for
the activity of a pure liquid or solid
a i ≈ 1 (pure liquid or solid near P
◦ )
(7.2-3)
The activities of pure liquids or solids can be omitted from an activity quotient unless
the pressure is very large.
E X A M P L E 7.4
a. Write the equilibrium constant quotient for the reaction CaCO 3 (s) CaO(s) + CO 2 (g).
b. Find the value of the equilibrium constant at 298.15 K and P eq (CO 2 ) at 298.15 K.
Solution
a. Since the activities of the solids are nearly equal to unity,
K
a eq (CaO)a eq (CO 2 )
a eq (CaCO 3 )
≈ a eq (CO 2 ) ≈
P eq (CO 2 )
P ◦
b. From the Gibbs energy changes of formation
∆G ◦ (1)(−603.501 kJ mol −1 ) − 394.389 kJ mol −1
+ (−1)(−1128.79 kJ mol −1 )
130.90 kJ mol −1