Benene diazonium chloride in aqueous solution decomposes as : `C_6H_5-N=N^(+)Cl_(aq)^(-) +H_2O_(aq)rarrC_6H_5OH_(aq)+N_2(g)+HCl_(aq)` The reaction follows first order kinetics. If `P_t` is the pressure of `N_2` at time t and `P_f` that after completion of the reaction, at constant volume and temperature, then which of the following graphs conforms to the kinetics of the reaction ?

Options

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Answer

To solve the problem regarding the decomposition of benzene diazonium chloride and its kinetics, we will follow these steps:

### Step 1: Understand the Reaction
The given reaction is:
\[ C_6H_5-N=N^+(Cl^-) + H_2O \rightarrow C_6H_5OH + N_2 + HCl \]

This reaction indicates that benzene diazonium chloride decomposes in the presence of water to produce phenol, nitrogen gas, and hydrochloric acid.

### Step 2: Identify the Kinetics
The problem states that this reaction follows first-order kinetics. For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant.

### Step 3: Write the Integrated Rate Law
The integrated rate law for a first-order reaction can be expressed as:
\[ k = \frac{2.303}{t} \log \left( \frac{a}{a - x} \right) \]
Where:
- \( k \) is the rate constant,
- \( t \) is the time,
- \( a \) is the initial concentration,
- \( x \) is the amount reacted.

### Step 4: Relate Pressure to Concentration
In this case, we can relate the pressure of nitrogen gas produced to the concentration of the reactant. Let:
- \( P_t \) be the pressure of nitrogen at time \( t \),
- \( P_f \) be the pressure of nitrogen at the completion of the reaction.

The initial pressure \( P_f \) corresponds to the initial concentration of benzene diazonium chloride. As the reaction progresses, the pressure \( P_t \) increases as nitrogen gas is produced.

### Step 5: Substitute into the Rate Law
We can express the concentrations in terms of pressures:
- \( a \) is proportional to \( P_f \),
- \( a - x \) is proportional to \( P_f - P_t \).

Thus, we can rewrite the integrated rate law:
\[ k = \frac{2.303}{t} \log \left( \frac{P_f}{P_f - P_t} \right) \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ \log (P_f - P_t) = -\frac{k}{2.303} t + \log P_f \]

This equation is in the form of \( y = mx + b \), where:
- \( y = \log (P_f - P_t) \),
- \( m = -\frac{k}{2.303} \) (the slope),
- \( x = t \),
- \( b = \log P_f \) (the y-intercept).

### Step 7: Graph Interpretation
The graph of \( \log (P_f - P_t) \) versus time \( t \) will be a straight line with a negative slope, indicating that as time increases, the logarithm of the pressure difference decreases.

### Conclusion
The correct graph that represents the kinetics of the reaction will show a straight line with a negative slope when plotting \( \log (P_f - P_t) \) against time \( t \).