Predict the entropy in the following change :
$H_{2}(g)\rightarrow2H(g)$

Question

Predict the entropy in the following change : 
$H_{2}(g)\rightarrow2H(g)$

Accepted Answer

In the given chemical change, $H_{2}(g) \rightarrow 2H(g)$, we are observing the dissociation of a diatomic hydrogen molecule into two individual hydrogen atoms. Let's analyze why the entropy increases in this process, supporting Option 3 as the correct answer.

### Explanation for Option 3: Increases

1. **Entropy and Molecular Disorder**: Entropy is a measure of the disorder or randomness in a system. When a diatomic molecule like $H_{2}$ dissociates into two separate atoms, the number of particles in the system increases. This increase in the number of particles leads to a greater number of possible microstates or configurations that the system can adopt, thereby increasing the disorder.

2. **Freedom of Movement**: In the initial state, the two hydrogen atoms are bound together in a single molecule, which restricts their movement relative to each other. Upon dissociation, each hydrogen atom can move independently, increasing the degrees of freedom and thus the entropy of the system.

3. **Statistical Mechanics Perspective**: From a statistical mechanics viewpoint, the entropy is related to the number of ways the particles can be arranged. The dissociation of $H_{2}$ into two separate $H$ atoms increases the number of possible arrangements, leading to an increase in entropy.

### Explanation for Why Other Options are Incorrect

1. **Option 1: 0**: This option suggests that there is no change in entropy. However, as explained, the dissociation process increases the number of particles and possible configurations, which inherently increases entropy. Therefore, the entropy change cannot be zero.

2. **Option 2: Remains same**: This option implies that the entropy before and after the reaction is identical. Given the increase in the number of particles and the associated increase in disorder, the entropy cannot remain the same. The system becomes more disordered, leading to an increase in entropy.

3. **Option 4: Decreases**: This option suggests a decrease in entropy, which contradicts the fundamental principle that breaking a molecule into more particles increases disorder. The dissociation process increases the number of independent particles, thus increasing the entropy, not decreasing it.

In conclusion, the dissociation of $H_{2}(g)$ into $2H(g)$ results in an increase in entropy due to the increase in the number of particles and the associated increase in disorder and possible configurations. Therefore, Option 3 is the correct answer.