The standard molar entropy of $H_{2}O_{(l)}$ is $70 \ J \ K^{-1} \ mol^{-1}$. Will the standard molar entropy of $H_{2}O_{(s)}$ be more or less than $70 \ J \ K^{-1} \ mol^{-1}$?
(N/A) The standard molar entropy of $H_{2}O_{(l)}$ is $70 \ J \ K^{-1} \ mol^{-1}$.
The solid form of $H_{2}O$ is ice. In ice,molecules of $H_{2}O$ are held in a rigid crystalline structure,making them less random and disordered than in liquid water.
Entropy is a measure of the randomness or disorder of a system. Since the solid state has a more ordered arrangement of molecules compared to the liquid state,the entropy of the solid is lower.
Therefore,the standard molar entropy of $H_{2}O_{(s)}$ will be less than $70 \ J \ K^{-1} \ mol^{-1}$.
The solid form of $H_{2}O$ is ice. In ice,molecules of $H_{2}O$ are held in a rigid crystalline structure,making them less random and disordered than in liquid water.
Entropy is a measure of the randomness or disorder of a system. Since the solid state has a more ordered arrangement of molecules compared to the liquid state,the entropy of the solid is lower.
Therefore,the standard molar entropy of $H_{2}O_{(s)}$ will be less than $70 \ J \ K^{-1} \ mol^{-1}$.
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The $S^o$ of the following substances are:
$CH_{4(g)} : 186.2 \ J \ K^{-1} \ mol^{-1}$
$O_{2(g)} : 205.2 \ J \ K^{-1} \ mol^{-1}$
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$H_2O_{(l)} : 69.9 \ J \ K^{-1} \ mol^{-1}$
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$H_2O_{(l)} : 69.9 \ J \ K^{-1} \ mol^{-1}$
The entropy change $\Delta S^o$ . . . . . . $J \ K^{-1} \ mol^{-1}$ for the reaction
$CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2H_2O_{(l)}$ is
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