A reaction `A + B toC + D + q` is found to have a positive entropy change. The reaction will be

Options

A. possible at high temperature
B. possible only at low temperature
C. not possible at any temperature
D. possible at any temperature.

Explanation

To solve the question regarding the reaction \( A + B \rightarrow C + D + q \) with a positive entropy change, we can analyze the thermodynamic principles involved. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the nature of the reaction
The reaction is given as \( A + B \rightarrow C + D + q \). The term \( + q \) indicates that heat is released during the reaction, which means that it is an **exothermic reaction**.

**Hint:** Remember that exothermic reactions release heat, which is indicated by the positive \( q \) in the equation.

### Step 2: Understand the implications of a positive entropy change
The problem states that the reaction has a **positive entropy change** (\( \Delta S > 0 \)). This means that the disorder of the system increases as the reaction proceeds.

**Hint:** A positive change in entropy usually indicates that the products have more freedom of movement or greater randomness compared to the reactants.

### Step 3: Apply the Gibbs Free Energy equation
We can use the Gibbs Free Energy equation:
\[ \Delta G = \Delta H - T \Delta S \]
Where:
- \( \Delta G \) is the change in Gibbs Free Energy,
- \( \Delta H \) is the change in enthalpy,
- \( T \) is the temperature in Kelvin,
- \( \Delta S \) is the change in entropy.

Since the reaction is exothermic, \( \Delta H \) is negative. Given that \( \Delta S \) is positive, we can analyze the sign of \( \Delta G \).

**Hint:** Remember that for a reaction to be spontaneous, \( \Delta G \) must be negative.

### Step 4: Analyze the signs of \( \Delta H \) and \( \Delta S \)
- \( \Delta H < 0 \) (negative for exothermic reactions)
- \( \Delta S > 0 \) (positive as given)

Substituting these into the Gibbs Free Energy equation:
\[ \Delta G = \text{(negative)} - T \times \text{(positive)} \]
This indicates that \( \Delta G \) will always be negative at any temperature because the negative enthalpy term dominates.

**Hint:** The temperature \( T \) is always positive, so multiplying it by a positive \( \Delta S \) will not change the fact that \( \Delta H \) is negative.

### Step 5: Conclusion
Since \( \Delta G \) is negative at all temperatures, the reaction is spontaneous and possible at any temperature.

**Final Answer:** The reaction is **possible at any temperature**.