(D) The entropy change for the reaction is calculated using the formula: $\Delta S^o = \sum \nu_p S^o_p - \sum \nu_r S^o_r$. For the reaction $H_{2(g)

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(D) The entropy change for the reaction is calculated using the formula: $\Delta S^o = \sum \nu_p S^o_p - \sum \nu_r S^o_r$. For the reaction $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}$,the expression is: $\Delta S^o = 2S^o(HCl) - [S^o(H_2) + S^o(Cl_2)]$. Substituting the given values: $\Delta S^o = 2 \times 186.7 - (130.6 + 223.0)$. $\Delta S^o = 373.4 - 353.6 = 19.8 \, J \, K^{-1} \, mol^{-1}$.

Class 11 Chemistry Thermodynamics 2 nd Law of thermodynamics and Entropy

Medium

The entropy values (in $J \, K^{-1} \, mol^{-1}$) of $H_{2(g)} = 130.6$,$Cl_{2(g)} = 223.0$,and $HCl_{(g)} = 186.7$ at $298 \, K$ and $1 \, atm$ pressure. Then entropy change for the reaction $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}$ is

A
$540.3$
B
$727.3$
C
$-166.9$
D
$19.8$

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Thermodynamics

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2 nd Law of thermodynamics and Entropy

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The entropy values (in $J \, K^{-1} \, mol^{-1}$) of $H_{2(g)} = 130.6$,$Cl_{2(g)} = 223.0$,and $HCl_{(g)} = 186.7$ at $298 \, K$ and $1 \, atm$ pressure. Then entropy change for the reaction $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}$ is

Similar Questions

During which of the following processes does entropy decrease?

The total amount of energy in the universe is fixed,but

Given $\Delta S_{Total} = -40 \ kJ/mol \cdot K$,$\Delta H_{System} = 2000 \ kJ/mol$,and $T = 400 \ K$. Find the value of $\Delta S_{System}$ in $kJ/mol \cdot K$.

At $373 \, K$,steam and water are in equilibrium and $\Delta H = 39.2 \, kJ \, mol^{-1}$. What will be $\Delta S$ for the conversion of $1 \, mole$ of water into steam?$H_2O_{(l)} \to H_2O_{(g)}$ ... $J \, K^{-1} \, mol^{-1}$

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At $373 \, K$,steam and water are in equilibrium and $\Delta H = 39.2 \, kJ \, mol^{-1}$. What will be $\Delta S$ for the conversion of $1 \, mole$ of water into steam?
$H_2O_{(l)} \to H_2O_{(g)}$ ... $J \, K^{-1} \, mol^{-1}$