Predict whether S for each reaction would be greater than zero, less than zero, or too close to zero to decide. Clear All H2 (g) + F2 (g) 2HF (g) 2NOBr (g) 2NO (g) + Br2 (g) CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) 2CO (g) + 2NO (g) 2CO2 (g) + N2 (g) 2H2O2 (l) 2H2O (l) + O2 (g) S > 0 S < 0 too close to decide Submit Answer

The symbol S represents Entropy. It is referred to as the degree of disorder or randomness in the system. In relation to phase changes;

Reactions involving solid - -> liquid - -> gas have higher (positive) entropy while;

Reactions involving gas - -> liquid - -> solid have lower (negative) values.

H2 (g) + F2 (g) - -> 2HF (g)

less than zero.

This is beacus the total number of entities has reduced from 2 gases to 1 gas, hence a lesser number of randomness.

2NOBr (g) - -> 2NO (g) + Br2 (g)

greater than zero

This is beacsue two entities were formed from one. Theres an increase in the disorderliness of the system.

CH4 (g) + 2O2 (g) - -> CO2 (g) + 2H2O (g)

too close to zero to decide. This is because the reactant and product side both have same number of entities.

2CO (g) + 2NO (g) - -> 2CO2 (g) + N2 (g)

too close to zero to decide. This is because the reactant and product side both have same number of entities.

2H2O2 (l) - -> 2H2O (l) + O2 (g)

Greater than Zero. Theres a phase change form liquid to gas. That accounts for the increase in disordeliness of the system.