For `PCl_(5)hArr PCl_(3)+Cl_(2)`, initial concentration of each reactant and product is 1 M. If `K_(eq)=0.41` then

Options

A. More `PCl_(3)` will form
B. More `Cl_(2)` will form
C. More `PCl_(5)` will form
D. No change

Explanation

To solve the problem regarding the equilibrium of the reaction \( PCl_5 \rightleftharpoons PCl_3 + Cl_2 \) with an initial concentration of 1 M for each species and an equilibrium constant \( K_{eq} = 0.41 \), we can follow these steps:

### Step 1: Write the equilibrium expression
The equilibrium constant expression for the reaction is given by:

\[ K_{eq} = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]

### Step 2: Set up the initial concentrations
Initially, we have:
- \([PCl_5] = 1 \, M\)
- \([PCl_3] = 1 \, M\)
- \([Cl_2] = 1 \, M\)

### Step 3: Define the change in concentrations
Let \( x \) be the change in concentration of \( PCl_5 \) that reacts to reach equilibrium. Therefore, at equilibrium, the concentrations will be:
- \([PCl_5] = 1 - x\)
- \([PCl_3] = 1 + x\)
- \([Cl_2] = 1 + x\)

### Step 4: Substitute into the equilibrium expression
Substituting these equilibrium concentrations into the equilibrium expression gives:

\[ K_{eq} = \frac{(1 + x)(1 + x)}{(1 - x)} = 0.41 \]

### Step 5: Simplify the equation
This simplifies to:

\[ K_{eq} = \frac{(1 + x)^2}{(1 - x)} = 0.41 \]

### Step 6: Cross multiply and rearrange
Cross multiplying gives:

\[ (1 + x)^2 = 0.41(1 - x) \]

Expanding both sides:

\[ 1 + 2x + x^2 = 0.41 - 0.41x \]

Rearranging to form a quadratic equation:

\[ x^2 + (2 + 0.41)x + (1 - 0.41) = 0 \]

This simplifies to:

\[ x^2 + 2.41x + 0.59 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 1 \), \( b = 2.41 \), and \( c = 0.59 \).

Calculating the discriminant:

\[ D = b^2 - 4ac = (2.41)^2 - 4(1)(0.59) = 5.8081 - 2.36 = 3.4481 \]

Now, calculate \( x \):

\[ x = \frac{-2.41 \pm \sqrt{3.4481}}{2} = \frac{-2.41 \pm 1.857}{2} \]

Calculating the two possible values for \( x \):

1. \( x = \frac{-2.41 + 1.857}{2} = \frac{-0.553}{2} = -0.2765 \) (not valid since concentration cannot be negative)
2. \( x = \frac{-2.41 - 1.857}{2} = \frac{-4.267}{2} = -2.1335 \) (also not valid)

### Step 8: Analyze the equilibrium
Since \( K_{eq} < 1 \), it indicates that the reaction favors the reactants. Therefore, more \( PCl_5 \) will form at equilibrium compared to the products \( PCl_3 \) and \( Cl_2 \).

### Conclusion
Thus, the correct answer is that more \( PCl_5 \) will form at equilibrium.