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]
