Consider the following equilibrium
`N_(2)O_(4)(g)hArr2NO_(2)(g)`
Then the select the correct graph , which shows the variation in concentrations of `N_(2)O_(4)` against concentrations of `NO_(2)`:

Options

A.
B.
C.
D.

Answer

To solve the problem of determining the correct graph that shows the variation in concentrations of \( N_2O_4 \) against concentrations of \( NO_2 \) for the equilibrium reaction:

\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

we will follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K_c \)

For the given equilibrium reaction, the equilibrium constant \( K_c \) is defined as:

\[ K_c = \frac{[NO_2]^2}{[N_2O_4]} \]

Where:
- \([NO_2]\) is the concentration of \( NO_2 \)
- \([N_2O_4]\) is the concentration of \( N_2O_4 \)

### Step 2: Rearrange the equation

From the expression for \( K_c \), we can rearrange it to express \([NO_2]^2\) in terms of \([N_2O_4]\):

\[ [NO_2]^2 = K_c \cdot [N_2O_4] \]

### Step 3: Define the variables

Let:
- \( Y = [NO_2] \)
- \( X = [N_2O_4] \)

Then we can rewrite the equation as:

\[ Y^2 = K_c \cdot X \]

### Step 4: Identify the type of graph

The equation \( Y^2 = K_c \cdot X \) represents a parabolic relationship between \( Y \) and \( X \). This means that as the concentration of \( N_2O_4 \) (X-axis) increases, the concentration of \( NO_2 \) (Y-axis) will increase in a quadratic manner.

### Step 5: Analyze the graph options

Given that we are looking for a parabolic graph, we can eliminate any linear graphs. The correct graph will be one that opens to the right, indicating that \( Y^2 \) increases as \( X \) increases.

### Conclusion

The correct graph that shows the variation in concentrations of \( N_2O_4 \) against concentrations of \( NO_2 \) is graph B, which represents a parabolic relationship.

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