Question

The standard entropies of \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\] (g), C (s) and \[{{\rm{O}}_{\rm{2}}}\] (g) are 213.5, 5.690 and 205 J/K respectively. The standard entropy of formation of \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\](g) is?
A. 1.86 J/K
B. 1.96 J/K
C. 2.81 J/K
D. 2.86 J/K

Answer 1

Hint: 2^nd law of thermodynamics says that, there is an entropy increase in the universe due to the change of entropy. The mathematical representation for the law is,
\[\Delta {S_{total}} = \Delta {S_{system}} + \Delta {S_{surrounding}} > 0\]

Complete step by step answer:
Let's understand what the entropy of formation is. The difference of standard entropy of products and the entropy of reactants is termed standard entropy of formation. It is symbolically represented as \[\Delta {S^0}\] . So, the formula to calculate entropy of formation is,
\[\Delta {S^o} = {S_{reactant}} - {S_{product}}\]

Let's come to the question. Here, carbon undergoes a reaction with oxygen to give carbon dioxide. The chemical reaction is,
\[C(s) + {{\rm{O}}_{\rm{2}}}(g) \to {\rm{C}}{{\rm{O}}_{\rm{2}}}(g)\]
Given the entropies of \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\] (g), is 213.5 J/K, C (s) is 5.690 J/K and \[{{\rm{O}}_{\rm{2}}}\] (g) is 205 J/K respectively.
So, the value of \[\Delta {S^0}\]is,
 \[\Delta {S^0} = 213.5 - 5.690 - 205 = 2.81\,{\rm{J/K}}\]
Therefore, entropy of \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\]formation is 2.81 J/K.

Hence, the right answer is option C.

The 1^st law of thermodynamics says the difference of the addition of heat to the system and work completed by the system gives the system's internal energy change. The mathematical representation is,
\[\Delta U = Q - W\]
Here, Q is heat and W is work and \[\Delta U\]is internal energy change.

Note: It is to be noted that, Positive value of \[\Delta S\] indicates that the proceeding of the reaction towards a state which possesses a greater degree of disorder and the negative value of \[\Delta S\]indicates the proceeding of the reaction towards that state which is more ordered.