The equilibrium constant, Kc, for the following reaction is 1.0 × 10^-5 at 1500k. N2(g) + O2(g) = 2NO(g) If 0.570 M N2 and 0.570 M O2 are allowed to equilibrate at 1500 K, what is the concentration of NO?
The equilibrium constant, Kc, for the following reaction is 1.0 × 10^-5 at 1500k. N2(g) + O2(g) = 2NO(g) If 0.570 M N2 and 0.570 M O2 are allowed to equilibrate at 1500 K, what is the concentration of NO?
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The equilibrium constant, Kc, for the following reaction is 1.0 × 10^-5 at 1500k.
N2(g) + O2(g) = 2NO(g)
If 0.570 M N2 and 0.570 M O2 are allowed to equilibrate at 1500 K, what is the concentration of NO?
![5. Consider the reaction A(aq) 2 B(aq) where K. 4.1 at 25 "C. If 0,50M A(aq) and1.5 M B(sq) are
CHEM 102
Equilibrium Study Guide
initially present in a 1.0L lask at 25 C. what change in concentrations (if any) will occu
[A] will decrease and [B] will decrease.
[A] will decrease and [B) will increase.
[A] will increase and [B] will decrease.
d.
a.
76
c.
(A] will increase and [B) will increase.
[A] and [B] remain unchanged.
Lo.so 9
2VKS
e.
0. An aqueous mixture of hydrocyanic acid and ammonia has initial concentrations of 0.100 MM ICNRD
0.140 M NH3(aq). At equilibrium, the CN (a0) concentration is 0.055 M. Calculate K for the reacto
HCN(aq) + NH:(aq) = CN (aq) + NH,"(aq)
a. 0.22
b. 0,79
c. 1.5
d.
3.9
e.
14
7. Nitrosyl bromide decomposes according to the chemical equation below.
2 NOBR(g) = 2 NO(g) + Br:(g)
When 0.260 atm of NOBR is sealed in a flask and allowed to reach equilibrium, 22% of the NOBr
decomposes. What is the equilibrium constant, Kp, for the reaction?
2.3 x 10
b. 4.5 x 10-3
3.5 x 10-2
d. 4.8 x 102
8.0 x 10-2
a.
с.
e.
8. The equilibrium constant, Ke, for the following reaction is 1.0 x 10 S at 1500 K.
N2(g) + O2(g) =2 NO(g)
If 0.570 M N2 and 0.570 M O2 are allowed to equilibrate at 1500 K, what is the concentration of NO?
5.7 x 10-6 M
a.
b. 9.0 x 104M
c. 2.4 x 10M
d. 1.8 x 10-3 M
2.4 x 10-3 M
e.
9. The equilibrium constant (K.) for the following reaction is 6.7 x 10-10 at 630 °C.
N2(s) + O2(g) =2 NO(g)
What is the equilibrium constant for the reaction below at the same temperature?
NO(g) = 1/2 N2(g) + 1/2 O2(g)
3.9 x 104
a.
b. 5.5 x 104
7.5 x 108
с.
d. 1.5 x 10°
e. 3.0 x 10°](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F325a91f3-8d85-4e09-afe5-577f0476af29%2Fd34777ed-1a5d-4b3f-9c91-2ce03df58df1%2Fdnl8l8r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:5. Consider the reaction A(aq) 2 B(aq) where K. 4.1 at 25 "C. If 0,50M A(aq) and1.5 M B(sq) are
CHEM 102
Equilibrium Study Guide
initially present in a 1.0L lask at 25 C. what change in concentrations (if any) will occu
[A] will decrease and [B] will decrease.
[A] will decrease and [B) will increase.
[A] will increase and [B] will decrease.
d.
a.
76
c.
(A] will increase and [B) will increase.
[A] and [B] remain unchanged.
Lo.so 9
2VKS
e.
0. An aqueous mixture of hydrocyanic acid and ammonia has initial concentrations of 0.100 MM ICNRD
0.140 M NH3(aq). At equilibrium, the CN (a0) concentration is 0.055 M. Calculate K for the reacto
HCN(aq) + NH:(aq) = CN (aq) + NH,"(aq)
a. 0.22
b. 0,79
c. 1.5
d.
3.9
e.
14
7. Nitrosyl bromide decomposes according to the chemical equation below.
2 NOBR(g) = 2 NO(g) + Br:(g)
When 0.260 atm of NOBR is sealed in a flask and allowed to reach equilibrium, 22% of the NOBr
decomposes. What is the equilibrium constant, Kp, for the reaction?
2.3 x 10
b. 4.5 x 10-3
3.5 x 10-2
d. 4.8 x 102
8.0 x 10-2
a.
с.
e.
8. The equilibrium constant, Ke, for the following reaction is 1.0 x 10 S at 1500 K.
N2(g) + O2(g) =2 NO(g)
If 0.570 M N2 and 0.570 M O2 are allowed to equilibrate at 1500 K, what is the concentration of NO?
5.7 x 10-6 M
a.
b. 9.0 x 104M
c. 2.4 x 10M
d. 1.8 x 10-3 M
2.4 x 10-3 M
e.
9. The equilibrium constant (K.) for the following reaction is 6.7 x 10-10 at 630 °C.
N2(s) + O2(g) =2 NO(g)
What is the equilibrium constant for the reaction below at the same temperature?
NO(g) = 1/2 N2(g) + 1/2 O2(g)
3.9 x 104
a.
b. 5.5 x 104
7.5 x 108
с.
d. 1.5 x 10°
e. 3.0 x 10°
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