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

S represents Entropy in a system and it is defined as the degree of

disorderliness in a system. In relation to phase changes;

Reactions which involve conversion of solid to liquid or from liquid to gas usually have a higher entropy.

Reactions which involve conversion of gas to liquids and liquids to solids usually have a lower entropy.

This is less than zero because the total number of products formed reduced from 2 gases to 1 gas, hence a lesser degree of disorderliness.

This is greater than zero because two products were formed from one. The number of gas products increased from one to two thereby an increasing the degree of disorderliness of the system.

This is too close to zero to decide as a result of  the reactant and product having equal number of products with same state of matter.

This is too close to zero to decide as a result of  the reactant and product having equal number of products with same state of matter.

This is greater than zero because two products were formed from one. The number of gas products increased from one to two thereby an increasing the degree of randomness of the system.

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Explanation:

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.