Question

Predict whether ΔS for each reaction would be greater than zero, less than zero, or too close to zero to decide:

1. I2(g) + Cl2(g) → 2ICl
2. 2NH3(g) →N2(g) + 3H2(g)
3. CO2(g) + H2(g) → CO(g) + H2O(g)
4. 2CO(g) + 2NO(g) → 2CO2(g) + N2
(g) 5. 2H2O2(l) →2H2O(l) + O2(g)

Answer 1

Answer: 1).
`I_(2)(g) + Cl_(2)(g) \rightarrow 2ICl(g)` has
`\Delta S = 0`.

2).
`2NH_(3)(g) \rightarrow N_(2)(g) + 3H_(2)(g)` has
`\Delta S > 0`.

3).
`CO_(2)(g) + H_(2)(g) \rightarrow CO(g) + H_(2)O(g)` has
`\Delta S = 0`.

4).
`2CO(g) + 2NO(g) \rightarrow 2CO_(2)(g) + N_(2)(g)` has
`\Delta S < 0`.

5).
`2H_(2)O_(2)(l) \rightarrow 2H_(2)O(l) + O_(2)(g)` has
`\Delta S > 0`.

Step-by-step explanation:

Entropy is the increase in degree of randomness of a substance. When a solid is converted into liquid then disorderness among the molecules of a substance increases.

As a result, entropy increases when a solid state converts into a liquid state.

Similarly, when a liquid state converts into gaseous state then molecules move more randomly from one place to another. Hence, entropy further increases when solid or liquid state of substance changes into gaseous state.

1).
`I_(2)(g) + Cl_(2)(g) \rightarrow 2ICl(g)`

This shows that 2 gaseous reactants are giving 2 gaseous product. So, entropy will remain the same. Hence,
`\Delta S = 0` for this reaction.

2).
`2NH_(3)(g) \rightarrow N_(2)(g) + 3H_(2)(g)`

Here, 2 moles of ammonia gas is giving 4 moles of gaseous products. This means that number of gaseous moles are increasing. So, entropy will increase and therefore,
`\Delta S > 0`.

3).
`CO_(2)(g) + H_(2)(g) \rightarrow CO(g) + H_(2)O(g)`

Here, 2 moles of gaseous reactants are giving 2 moles of gaseous products . So, there is no increase in number of gaseous moles therefore,
`\Delta S = 0`.

4).
`2CO(g) + 2NO(g) \rightarrow 2CO_(2)(g) + N_(2)(g)`

Here, total 4 moles of gaseous reactants are giving total 3 moles of gaseous products. This means a decrease in number of moles is taking place. Hence, entropy is decreasing and therefore,
`\Delta S < 0`.

5).
`2H_(2)O_(2)(l) \rightarrow 2H_(2)O(l) + O_(2)(g)`

Here, 2 moles of reactants are giving total 3 moles of products. This means that an increase in number of moles is taking place. So, entropy will increase therefore,
`\Delta S > 0`.