The `K_c` for the reaction `A+B Leftrightarrow C` is 4 and K. for `2A+D Leftrightarrow C` 6. The value of `K_c" for "C+D Leftrightarrow 2B` is

Options

A. 0.67
B. 0.375
C. 2.7
D. 1.5

Answer

To find the equilibrium constant \( K_c \) for the reaction \( C + D \leftrightarrow 2B \), we will use the given equilibrium constants for the reactions:

1. \( A + B \leftrightarrow C \) with \( K_1 = 4 \)
2. \( 2A + D \leftrightarrow C \) with \( K_2 = 6 \)

### Step-by-Step Solution:

**Step 1: Manipulate the first reaction.**

We start with the first reaction:

\[ A + B \leftrightarrow C \quad (K_1 = 4) \]

We will multiply the entire reaction by 2:

\[ 2A + 2B \leftrightarrow 2C \]

When we multiply a reaction by a coefficient, we raise the equilibrium constant to the power of that coefficient. Therefore, the new equilibrium constant \( K_1' \) will be:

\[ K_1' = K_1^2 = 4^2 = 16 \]

**Step 2: Reverse the modified first reaction.**

Now, we reverse the reaction:

\[ 2C \leftrightarrow 2A + 2B \]

When we reverse a reaction, the equilibrium constant becomes the reciprocal of the original. Thus, the new equilibrium constant \( K_1'' \) is:

\[ K_1'' = \frac{1}{K_1'} = \frac{1}{16} \]

**Step 3: Add the second reaction.**

Now we add the second reaction:

\[ 2A + D \leftrightarrow C \quad (K_2 = 6) \]

Adding the two reactions together:

\[ 2C \leftrightarrow 2A + 2B \quad (K_1'') \\ 2A + D \leftrightarrow C \quad (K_2) \]

The overall reaction becomes:

\[ C + D \leftrightarrow 2B \]

**Step 4: Calculate the equilibrium constant for the overall reaction.**

When we add two reactions, the equilibrium constants multiply:

\[ K_c = K_1'' \times K_2 = \left(\frac{1}{16}\right) \times 6 \]

Calculating this gives:

\[ K_c = \frac{6}{16} = \frac{3}{8} = 0.375 \]

### Final Answer:

The value of \( K_c \) for the reaction \( C + D \leftrightarrow 2B \) is \( 0.375 \).

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