
The standard entropies of\[{{N}_{2}}(g)\], \[{{H}_{2}}(g)\] and \[N{{H}_{3}}(g)\] are 191.5, 130.5, 192.6 \[J{{K}^{-1}}mo{{l}^{-1}}\]. The value of \[\Delta {{S}^{o}}\]during the formation of one mole of ammonia is:
Answer
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Hint: The formation of ammonia from nitrogen and hydrogen is as follows.
\[\dfrac{1}{2}{{N}_{2(g)}}+\dfrac{3}{2}{{H}_{2(g)}}\to N{{H}_{3(g)}}\]
0.5 moles of nitrogen reacts with 1.5 moles of hydrogen and forms one mole of ammonia is a product.
Complete step by step answer:
- In the question it is given that the standard entropy of \[{{N}_{2}}(g)\](reactant) is191.5\[J{{K}^{-1}}mo{{l}^{-1}}\]
-The standard entropy of \[{{H}_{2}}(g)\](reactant) is 130.5\[J{{K}^{-1}}mo{{l}^{-1}}\]
-The standard entropy of \[N{{H}_{3}}(g)\](product) is 192.6 \[J{{K}^{-1}}mo{{l}^{-1}}\]
-There is a formula to calculate the change in standard entropy of a reaction. It is as follows.
\[\Delta {{S}^{o}}=\Delta {{S}^{o}}_{(product)}-\Delta {{S}^{o}}_{(reac\tan t)}\]
-Substitute all the known values in the above equation to get standard entropy of a reaction.
\[\begin{align}
& \Delta {{S}^{o}}=\Delta {{S}^{o}}_{(product)}-\Delta {{S}^{o}}_{(reac\tan t)} \\
& \Delta {{S}^{o}}=192.6-\left( \dfrac{1}{2}\times 191.5+\dfrac{3}{2}\times 130.5 \right) \\
& \Delta {{S}^{o}}=192.6-291.5 \\
& \Delta {{S}^{o}}=-98.9J{{K}^{-1}}mo{{l}^{-1}} \\
\end{align}\]
- The standard entropy of formation of ammonia in the reaction is -98.9 \[J{{K}^{-1}}mo{{l}^{-1}}\]
- So, the value of \[\Delta {{S}^{o}}\]during the formation of one mole of ammonia is -98.9 \[J{{K}^{-1}}mo{{l}^{-1}}\]
Note: The standard entropy is the entropy of one mole of pure substance or compound under a standard state. The standard entropy is going to be denoted with a symbol \[\Delta {{S}^{o}}\]with the units\[J{{K}^{-1}}mo{{l}^{-1}}\].
Standard entropy of the reaction is the subtraction of standard entropy of reactants from standard entropy of products with respect to their number moles involved in the reaction.
Standard entropy is nothing but the degree of randomness of one mole of the reactants or products at standard reaction conditions. A positive value of Standard entropy specifies an upsurge in entropy of the reaction and a negative value specifies a reduction in the entropy of the reaction.
\[\dfrac{1}{2}{{N}_{2(g)}}+\dfrac{3}{2}{{H}_{2(g)}}\to N{{H}_{3(g)}}\]
0.5 moles of nitrogen reacts with 1.5 moles of hydrogen and forms one mole of ammonia is a product.
Complete step by step answer:
- In the question it is given that the standard entropy of \[{{N}_{2}}(g)\](reactant) is191.5\[J{{K}^{-1}}mo{{l}^{-1}}\]
-The standard entropy of \[{{H}_{2}}(g)\](reactant) is 130.5\[J{{K}^{-1}}mo{{l}^{-1}}\]
-The standard entropy of \[N{{H}_{3}}(g)\](product) is 192.6 \[J{{K}^{-1}}mo{{l}^{-1}}\]
-There is a formula to calculate the change in standard entropy of a reaction. It is as follows.
\[\Delta {{S}^{o}}=\Delta {{S}^{o}}_{(product)}-\Delta {{S}^{o}}_{(reac\tan t)}\]
-Substitute all the known values in the above equation to get standard entropy of a reaction.
\[\begin{align}
& \Delta {{S}^{o}}=\Delta {{S}^{o}}_{(product)}-\Delta {{S}^{o}}_{(reac\tan t)} \\
& \Delta {{S}^{o}}=192.6-\left( \dfrac{1}{2}\times 191.5+\dfrac{3}{2}\times 130.5 \right) \\
& \Delta {{S}^{o}}=192.6-291.5 \\
& \Delta {{S}^{o}}=-98.9J{{K}^{-1}}mo{{l}^{-1}} \\
\end{align}\]
- The standard entropy of formation of ammonia in the reaction is -98.9 \[J{{K}^{-1}}mo{{l}^{-1}}\]
- So, the value of \[\Delta {{S}^{o}}\]during the formation of one mole of ammonia is -98.9 \[J{{K}^{-1}}mo{{l}^{-1}}\]
Note: The standard entropy is the entropy of one mole of pure substance or compound under a standard state. The standard entropy is going to be denoted with a symbol \[\Delta {{S}^{o}}\]with the units\[J{{K}^{-1}}mo{{l}^{-1}}\].
Standard entropy of the reaction is the subtraction of standard entropy of reactants from standard entropy of products with respect to their number moles involved in the reaction.
Standard entropy is nothing but the degree of randomness of one mole of the reactants or products at standard reaction conditions. A positive value of Standard entropy specifies an upsurge in entropy of the reaction and a negative value specifies a reduction in the entropy of the reaction.
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