Change in a chemical reaction

Question: which of the following options are correct for the given equation?

    

a)      Phase transition

b)      Allotropic change

c)       ∆H is positive

d)       ∆S is positive 

e)      ∆S is negative   

Solution: Let us observe each option with the given chemical reaction carefully.

a)      Phase Transition:

In the given reaction, solid carbon-dioxide changes into gaseous form. This gives clear hint that option (a) is correct for the equation because there is a phase transition. If you remembered the basic than you should know that a phase transition is the transformation of a substance from one phase or state of matter to another. (here solid changes into gas)

b)      Allotropic Change

There is no allotropic change in the reaction because the term allotrope refers to one or more forms of an elementary substance. Thus, option (b) is wrong for the equation. (because Carbon dioxide is a compound)

c)       ∆H is positive

Delta H refers to change in heat in the reaction. So let us look our reaction again. 

In reaction solid carbon dioxide changes directly into gaseous form.  This type of transition is called sublimation which requires additional energy.  The system absorbs this additional energy from its surroundings in the form of heat. Such kind of reaction is known as endothermic reaction. 

The change ΔH is positive in endothermic reactions, and negative in heat-releasing exothermic processes.

Thus, option (c) is correct for the equation. 

d)      ∆S is positive 

Delta S refers to change in entropy. Entropy is a mathematically-defined thermodynamic quantity that helps to account for the flow of energy through a thermodynamic process.

Change in entropy is positive when....
-you go from a solid to a liquid
-you go from a liquid to a gas
-when you go from a solid to a gas
-when there are more mols of products than mols of reactant

And similarly change in entropy is negative when the reverse of the above happens.

In the above reaction, Solid CO₂ changes into gas, so ∆S must be positive here.

Thus, option (d) is correct for the equation. 

e)      ∆S is negative 

If option (d) s correct for the equation, then option (e) must be wrong.

Thus, options (a), (c) and (d) are correct for the equation.