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At equilibrium, the concentrations in this system were found to be [tex]\left[ N_2 \right] = \left[ O_2 \right] = 0.100 \, M[/tex] and [tex][ NO ] = 0.500 \, M[/tex].

[tex]\[ N_2(g) + O_2(g) \rightleftharpoons 2 \, NO(g) \][/tex]

If more NO is added, bringing its concentration to [tex]0.800 \, M[/tex], what will the final concentration of NO be after equilibrium is re-established?



Answer :

GinnyAnswer
Let's solve this step-by-step:

### Step 1: Given Data
We are given the following initial equilibrium concentrations:
- [tex]\([N_2]_{\text{initial}} = 0.100 \text{ M}\)[/tex]
- [tex]\([O_2]_{\text{initial}} = 0.100 \text{ M}\)[/tex]
- [tex]\([NO]_{\text{initial}} = 0.500 \text{ M}\)[/tex]

And new concentration of NO:
- [tex]\([NO]_{\text{added}} = 0.800 \text{ M}\)[/tex]

### Step 2: Determine the Equilibrium Constant ([tex]\(K_{eq}\)[/tex])

The reaction is:
[tex]\[ N_2(g) + O_2(g) \rightleftharpoons 2 NO(g) \][/tex]

The equilibrium constant expression for the reaction is:
[tex]\[ K_{eq} = \frac{[NO]^2}{[N_2][O_2]} \][/tex]

Using the initial equilibrium concentrations:
[tex]\[ K_{eq} = \frac{(0.500)^2}{(0.100)(0.100)} = \frac{0.250}{0.010} = 25 \][/tex]

### Step 3: Set Up the Change in Concentration

Let [tex]\(x\)[/tex] represent the change in concentration of [tex]\(NO\)[/tex] that will occur to re-establish equilibrium. Thus:
- The change in [tex]\([NO]\)[/tex] will be [tex]\(2x\)[/tex], as two moles of NO are involved for each mole of [tex]\(N_2\)[/tex] and [tex]\(O_2\)[/tex].
- Since [tex]\(N_2\)[/tex] and [tex]\(O_2\)[/tex] each react with one mole, their increase in concentration will each be [tex]\(x\)[/tex].

Thus, the concentrations at new equilibrium will be:
- [tex]\([NO] = 0.800 - 2x\)[/tex]
- [tex]\([N_2] = 0.100 + x\)[/tex]
- [tex]\([O_2] = 0.100 + x\)[/tex]

### Step 4: Set Up the Equilibrium Expression with the New Concentrations

Using the expression for [tex]\(K_{eq}\)[/tex]:
[tex]\[ K_{eq} = \frac{(0.800 - 2x)^2}{(0.100 + x)(0.100 + x)} \][/tex]

### Step 5: Substitute the Equilibrium Constant and Solve for x

We substitute [tex]\(K_{eq} = 25\)[/tex]:
[tex]\[ 25 = \frac{(0.800 - 2x)^2}{(0.100 + x)(0.100 + x)} \][/tex]

After solving this quadratic equation, we find the positive value of [tex]\(x\)[/tex]. The solution to this equation results in:
[tex]\[ x = 0.042857 \][/tex]

### Step 6: Calculate the Final Concentration of NO

Using the value of [tex]\(x\)[/tex], we can find the final concentration of NO:
[tex]\[ [NO]_{\text{final}} = 0.800 - 2x = 0.800 - 2(0.042857) = 0.800 - 0.085714 = 0.714286 \text{ M} \][/tex]

### Conclusion

The final concentration of NO after equilibrium is re-established is:
[tex]\[ [NO]_{\text{final}} = 0.714286 \text{ M} \][/tex]

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