At equilibrium, the concentrations in this system were found to be [N₂] = [0₂] = 0.100 M and [NO] N₂(g) + O₂(g) = 2 NO(g) If more NO is added, bringing its concentration to 0.900 M, what will the final concentration of NO be after equilibrium is re-established? = 0.600 M.
At equilibrium, the concentrations in this system were found to be [N₂] = [0₂] = 0.100 M and [NO] N₂(g) + O₂(g) = 2 NO(g) If more NO is added, bringing its concentration to 0.900 M, what will the final concentration of NO be after equilibrium is re-established? = 0.600 M.
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At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.100 M and [NO]=0.600 M.
N2(g) + O2(g)↽−−⇀ 2NO(g)
If more NO is added, bringing its concentration to 0.900 M, what will the final concentration of NO be after equilibrium is re‑established?
![### Understanding Chemical Equilibrium: A Case Study
**Problem Statement:**
At equilibrium, the concentrations in the following system were found to be \([N_2] = [O_2] = 0.100 \, \text{M}\) and \([NO] = 0.600 \, \text{M}\).
\[ \mathrm{N_2(g) + O_2(g) \rightleftharpoons 2NO(g)} \]
If additional NO is added, bringing its concentration to \(0.900 \, \text{M}\), what will be the final concentration of NO after equilibrium is re-established?
### Step-by-Step Solution:
1. **Initial Equilibrium Status:**
At the initial equilibrium, the following concentrations are given:
- \([N_2] = 0.100 \, \text{M}\)
- \([O_2] = 0.100 \, \text{M}\)
- \([NO] = 0.600 \, \text{M}\)
Using the equilibrium expression for the reaction:
\[ \mathrm{K_c = \frac{[NO]^2}{[N_2][O_2]}} \]
2. **Calculate the Equilibrium Constant (\(K_c\)):**
Substitute the given equilibrium concentrations into the equilibrium expression:
\[ \mathrm{K_c = \frac{(0.600)^2}{(0.100)(0.100)} = \frac{0.360}{0.010} = 36} \]
3. **Disturbance by Adding NO:**
When more NO is added to the system, bringing its concentration to \(0.900 \, \text{M}\):
New initial concentration of NO: \([NO]_{\text{initial}} = 0.900 \, \text{M}\)
4. **Re-establishing Equilibrium:**
Let's assume that \( \, x \, \text{M} \) of NO reacts to re-establish the equilibrium concentration.
The change in concentrations will be:
- NO decreases by \(2x\)
- \(N_2\) increases by \(x\)
- \(O_2\) increases by \(x\)
Therefore, at the new equilibrium:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85650e2f-17ad-4aa4-9d56-acd675fbc675%2F6a48939b-d752-48a0-bb3f-b9ca188885f7%2Fw3a22pd_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Chemical Equilibrium: A Case Study
**Problem Statement:**
At equilibrium, the concentrations in the following system were found to be \([N_2] = [O_2] = 0.100 \, \text{M}\) and \([NO] = 0.600 \, \text{M}\).
\[ \mathrm{N_2(g) + O_2(g) \rightleftharpoons 2NO(g)} \]
If additional NO is added, bringing its concentration to \(0.900 \, \text{M}\), what will be the final concentration of NO after equilibrium is re-established?
### Step-by-Step Solution:
1. **Initial Equilibrium Status:**
At the initial equilibrium, the following concentrations are given:
- \([N_2] = 0.100 \, \text{M}\)
- \([O_2] = 0.100 \, \text{M}\)
- \([NO] = 0.600 \, \text{M}\)
Using the equilibrium expression for the reaction:
\[ \mathrm{K_c = \frac{[NO]^2}{[N_2][O_2]}} \]
2. **Calculate the Equilibrium Constant (\(K_c\)):**
Substitute the given equilibrium concentrations into the equilibrium expression:
\[ \mathrm{K_c = \frac{(0.600)^2}{(0.100)(0.100)} = \frac{0.360}{0.010} = 36} \]
3. **Disturbance by Adding NO:**
When more NO is added to the system, bringing its concentration to \(0.900 \, \text{M}\):
New initial concentration of NO: \([NO]_{\text{initial}} = 0.900 \, \text{M}\)
4. **Re-establishing Equilibrium:**
Let's assume that \( \, x \, \text{M} \) of NO reacts to re-establish the equilibrium concentration.
The change in concentrations will be:
- NO decreases by \(2x\)
- \(N_2\) increases by \(x\)
- \(O_2\) increases by \(x\)
Therefore, at the new equilibrium:
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