9. What is Kc for: 2Cu(s) + O2(g)= 2CUO(s)? Given the following equilibrium and Kc's* 4Cu(s) + O2(g) = 2Cu,0(s) K = 1.7 x 10-4 2CuO(s) = Cu,0(s) + ½ O2(g) K = 4.3 x 102 O 0.073 O 120,000 O 3.0 x 10^-5 5.6 3.5 x 10^-3
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Chemical equilibrium and ionic equilibrium are two major concepts in chemistry. Ionic equilibrium deals with the equilibrium involved in an ionization process while chemical equilibrium deals with the equilibrium during a chemical change. Ionic equilibrium is established between the ions and unionized species in a system. Understanding the concept of ionic equilibrium is very important to answer the questions related to certain chemical reactions in chemistry.
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What is Kc for: 2 Cu (s) + O2 (g) ---> <--- 2CuO (s)? Given the following equilibrium and Kc's?
1. 0.073
2. 120,000
3. 3.0 x 10^-5
4. 5.6
5. 3.5 x 10^-3
![### Understanding the Calculation of Equilibrium Constants (Kc)
**Question 9:** What is Kc for the reaction:
\[ 2Cu(s) + O_2(g) \rightleftharpoons 2CuO(s) \]
Given the following equilibrium and Kc values:
1. \( 4Cu(s) + O_2(g) \rightleftharpoons 2Cu_2O(s) \)
\[ K_c = 1.7 \times 10^{-4} \]
2. \( 2CuO(s) \rightleftharpoons Cu_2O(s) + \frac{1}{2}O_2(g) \)
\[ K_c = 4.3 \times 10^2 \]
#### Provided Options:
- 0.073
- 120,000
- \( 3.0 \times 10^{-5} \)
- 5.6
- \( 3.5 \times 10^{-3} \)
This is a required question.
### Explanation of the Equilibrium Reaction
To determine the overall equilibrium constant \( K_c \) for the target reaction \( 2Cu(s) + O_2(g) \rightleftharpoons 2CuO(s) \), we need to manipulate the given reactions and their equilibrium constants appropriately.
1. The first given equilibrium reaction:
\[ 4Cu(s) + O_2(g) \rightleftharpoons 2Cu_2O(s), \, K_c = 1.7 \times 10^{-4} \]
2. The second given equilibrium reaction:
\[ 2CuO(s) \rightleftharpoons Cu_2O(s) + \frac{1}{2}O_2(g), \, K_c = 4.3 \times 10^2 \]
Reverse the direction of the second reaction to align with the target equation:
\[ Cu_2O(s) + \frac{1}{2}O_2(g) \rightleftharpoons 2CuO(s) \]
The equilibrium constant for this reversed reaction will be the reciprocal of the original Kc:
\[ K_c = \frac{1}{4.3 \times 10^2} \]
### Calculation:
1. Rewrite the first equation such that the desired substances appear on](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8466b748-e672-4c61-893b-8156cddfafb1%2Fb0099201-6c3d-4129-8141-32adce775c3e%2F0nqz9n7_processed.jpeg&w=3840&q=75)
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