In a flask colourless `N_2O_4` is in equilibrium with brown coloured `NO_2`. At equilibrium, when the flask is heated at `100^@C` the brown colour deepens and on cooling it becomes less coloured. The change in enthalpy, `DeltaH` for this system is

Options

A. negative
B. positive
C. zero
D. undefined

Explanation

To determine the change in enthalpy (ΔH) for the equilibrium between colorless \(N_2O_4\) and brown \(NO_2\), we can analyze the situation step-by-step.

### Step 1: Understand the Reaction
The equilibrium can be represented as:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]
In this reaction, \(N_2O_4\) is colorless, while \(NO_2\) is brown.

### Step 2: Analyze the Effect of Heating
When the flask is heated to \(100^\circ C\), the brown color deepens. This indicates that more \(NO_2\) is being produced, which means the equilibrium shifts to the right (towards the formation of \(NO_2\)).

### Step 3: Determine the Nature of the Reaction
The fact that heating the system increases the concentration of \(NO_2\) suggests that the reaction absorbs heat. This is characteristic of an endothermic reaction, where heat is a reactant.

### Step 4: Cooling the System
When the flask is cooled, the brown color becomes less intense, indicating that the concentration of \(NO_2\) decreases and the equilibrium shifts to the left (towards the formation of \(N_2O_4\)). This behavior is consistent with an endothermic reaction, as cooling would favor the reverse reaction.

### Step 5: Conclude the Change in Enthalpy
Since the reaction absorbs heat (endothermic), the change in enthalpy (ΔH) is positive:
\[ \Delta H > 0 \]

### Final Answer
The change in enthalpy, ΔH, for this system is positive.

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