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Solve for [tex]\( x \)[/tex]:
[tex]\[ 3x = 6x - 2 \][/tex]


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Question 2
A system at equilibrium in a 1.0-liter container was found to contain 0.20 mol of [tex]$A , 0.20 mol$[/tex] of [tex]$B , 0.40 mol$[/tex] of C , and 0.40 mol of D . If 0.15 mol of A and 0.15 mol of B are added to this system, what will be the new equilibrium concentration of A ?
[tex]$
A ( g )+ B ( g )\ \textless \ C ( g )+ D ( g )
$[/tex]
a. 0.050 M
b. 0.10 M
c. 0.20 M
d. 0.25 M
e. 0.30 M
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Response:

A system at equilibrium in a 1.0-liter container was found to contain 0.20 mol of [tex]\( A \)[/tex], 0.20 mol of [tex]\( B \)[/tex], 0.40 mol of [tex]\( C \)[/tex], and 0.40 mol of [tex]\( D \)[/tex]. If 0.15 mol of [tex]\( A \)[/tex] and 0.15 mol of [tex]\( B \)[/tex] are added to this system, what will be the new equilibrium concentration of [tex]\( A \)[/tex]?

[tex]\[ A(g) + B(g) \leftrightarrow C(g) + D(g) \][/tex]

a. 0.050 M
b. 0.10 M
c. 0.20 M
d. 0.25 M
e. 0.30 M



Answer :

GinnyAnswer
To solve this problem, we need to follow a systematic approach. We'll start by analyzing the initial state, adding the changes, using the equilibrium expression, and then finding the new equilibrium concentrations.

### Step 1: Initial State at Equilibrium
Initially, we have:
- [tex]\( [A]_0 = 0.20 \)[/tex] mol/L
- [tex]\( [B]_0 = 0.20 \)[/tex] mol/L
- [tex]\( [C]_0 = 0.40 \)[/tex] mol/L
- [tex]\( [D]_0 = 0.40 \)[/tex] mol/L

### Step 2: Adding A and B
We add 0.15 mol of A and 0.15 mol of B to the system:
- New concentration of A: [tex]\( [A]_0 + 0.15 = 0.20 + 0.15 = 0.35 \)[/tex] mol/L
- New concentration of B: [tex]\( [B]_0 + 0.15 = 0.20 + 0.15 = 0.35 \)[/tex] mol/L

Therefore, the concentrations immediately after adding A and B are:
- [tex]\( [A]_i = 0.35 \)[/tex] mol/L
- [tex]\( [B]_i = 0.35 \)[/tex] mol/L
- [tex]\( [C]_i = 0.40 \)[/tex] mol/L
- [tex]\( [D]_i = 0.40 \)[/tex] mol/L

### Step 3: Equilibrium Constant Calculation
The equilibrium constant [tex]\( K \)[/tex] is calculated based on the initial equilibrium concentrations:
[tex]\[ K = \frac{[C]_0[D]_0}{[A]_0[B]_0} = \frac{(0.40 \, \text{mol/L}) (0.40 \, \text{mol/L})}{(0.20 \, \text{mol/L}) (0.20 \, \text{mol/L})} = \frac{0.16}{0.04} = 4 \][/tex]

### Step 4: Setting Up the ICE Table
Now, we use the ICE (Initial, Change, Equilibrium) table to find the changes that occur as the reaction re-establishes equilibrium:
```
A (g) + B (g) <=> C (g) + D (g)
Initial: 0.35 0.35 0.40 0.40
Change: -x -x +x +x
Equilibrium: 0.35-x 0.35-x 0.40+x 0.40+x
```

### Step 5: Applying the Equilibrium Expression
At equilibrium:
[tex]\[ K = \frac{[C][D]}{[A][B]} \][/tex]
[tex]\[ 4 = \frac{(0.40 + x)(0.40 + x)}{(0.35 - x)(0.35 - x)} \][/tex]

### Step 6: Solving the Equilibrium Expression
We need to solve the equation:
[tex]\[ 4 = \frac{(0.40 + x)^2}{(0.35 - x)^2} \][/tex]
Taking the square root of both sides:
[tex]\[ 2 = \frac{0.40 + x}{0.35 - x} \][/tex]

Now, solve for [tex]\( x \)[/tex]:
[tex]\[ 2(0.35 - x) = 0.40 + x \][/tex]
[tex]\[ 0.70 - 2x = 0.40 + x \][/tex]
[tex]\[ 0.70 - 0.40 = 3x \][/tex]
[tex]\[ 0.30 = 3x \][/tex]
[tex]\[ x = \frac{0.30}{3} = 0.10 \][/tex]

### Step 7: New Equilibrium Concentration of A
Using the value of [tex]\( x \)[/tex] to find the new equilibrium concentration of A:
[tex]\[ [A]_{eq} = 0.35 - x = 0.35 - 0.10 = 0.25 \, \text{mol/L} \][/tex]

### Conclusion
Thus, the new equilibrium concentration of [tex]\( A \)[/tex] is:

[tex]\( \boxed{0.25 \, \text{mol/L}} \)[/tex]

So, the correct answer is [tex]\( \text{d.} \)[/tex]