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Question: If \({{S}^{0}}\) for \({{H}_{2}},C{{l}_{2}}\) and \(HCl\) are 0.13, 0.22 and 0.19 \(KJ{{K}^{-1}}mo{{...

If S0{{S}^{0}} for H2,Cl2{{H}_{2}},C{{l}_{2}} and HClHCl are 0.13, 0.22 and 0.19 KJK−1mol−1KJ{{K}^{-1}}mo{{l}^{-1}} respectively. The total change is standard entropy for the reaction, H2+Cl2→2HCl{{H}_{2}}+C{{l}_{2}}\to 2HCl is:
(A) 30 JK−1mol−1J{{K}^{-1}}mo{{l}^{-1}}
(B) 40 JK−1mol−1J{{K}^{-1}}mo{{l}^{-1}}
(C) 60 JK−1mol−1J{{K}^{-1}}mo{{l}^{-1}}
(D) 20 JK−1mol−1J{{K}^{-1}}mo{{l}^{-1}}

Explanation

Solution

The formulae to calculate total change in entropy is:
Totalchangeinentropy=entropyofproduct–entropyofreactantTotal change in entropy = entropy of product – entropy of reactant,
all the values are already provided in the question. Plugin that data and find the total change entropy.

Complete step by step solution:
Before solving the question let us understand what entropy means, entropy is the measure of the thermal energy of the system per unit temperature which is available for doing the work.
If we talk about work it is obtained from the ordered motion of the molecule whereas the amount of entropy is the measurement of disorder of a molecule, or we can say that randomness of the system.
Now from the above equation, we know that hydrogen and chlorine i.e. H2,Cl2{{H}_{2}}, C{{l}_{2}} are reactants whereas hydrochloric acid i.e. HCl is the product.
Given in the question the value of entropy
S0{{S}^{0}} For hydrogen H2{{H}_{2}} = 0.130.13 KJK−1mol−1KJ{{K}^{-1}}mo{{l}^{-1}}
S0{{S}^{0}} For chlorine Cl2C{{l}_{2}} = 0.130.13 KJK−1mol−1KJ{{K}^{-1}}mo{{l}^{-1}}
S0{{S}^{0}} For hydrochloric acid HCl = 0.190.19 KJK−1mol−1KJ{{K}^{-1}}mo{{l}^{-1}}
The total change in entropy is calculated by subtracting the total entropy of products by the total entropy of reactants.
Total change in entropy = entropy of product – entropy of reactant
ΔS0=∑S0products−∑S0reactan⁡t\Delta {{S}^{0}}={{\sum{{{S}^{0}}}}_{products}}-{{\sum{{{S}^{0}}}}_{reac\tan t}}
Put the given values of entropy of hydrogen, chlorine and hydrochloric acid in the above equation
ΔS0\Delta {{S}^{0}}= 2X0.19−(0.13+0.22)2X0.19-(0.13+0.22)
ΔS0\Delta {{S}^{0}}= 0.030.03 KJK−1mol−1KJ{{K}^{-1}}mo{{l}^{-1}}
ΔS0\Delta {{S}^{0}}= 3030 JK−1mol−1J{{K}^{-1}}mo{{l}^{-1}}
The total change in entropy is 3030 JK−1mol−1J{{K}^{-1}}mo{{l}^{-1}}

Hence, the correct option is option (A) which is 30 JK−1mol−1J{{K}^{-1}}mo{{l}^{-1}}.

Note: Be careful with the units, as in the question the value of entropy is given in kilojoule but in the options, the total change in entropy is given in joules. The conversion of kilojoules to joules is easy.
1 J = 0.001 KJ.