Questions: Calculate the standard entropy change for the reaction at (25^circ mathrmC). Standard molar entropy values can be found in this table. [ mathrmMg(mathrmOH)2(mathrm~s)+2 mathrmHCl(mathrm~g) longrightarrow mathrmMgCl2(mathrm~s)+2 mathrmH2 mathrmO(mathrm~g) Delta Smathrmixn=square ] (square) J/K TOOLS [ x 10^y ]

Solution Steps

The reaction given is:

\[ \mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{~s}) + 2 \mathrm{HCl}(\mathrm{~g}) \longrightarrow \mathrm{MgCl}_{2}(\mathrm{~s}) + 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \]

We need to calculate the standard entropy change (\(\Delta S^\circ\)) for this reaction at \(25^{\circ} \mathrm{C}\).

To calculate \(\Delta S^\circ\), we need the standard molar entropy values (\(S^\circ\)) for each substance involved in the reaction. These values are typically found in thermodynamic tables. Assume the following standard molar entropy values (in J/mol·K):

• \(S^\circ(\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})) = 62.77\)
• \(S^\circ(\mathrm{HCl}(\mathrm{g})) = 186.90\)
• \(S^\circ(\mathrm{MgCl}_{2}(\mathrm{s})) = 89.60\)
• \(S^\circ(\mathrm{H}_{2}\mathrm{O}(\mathrm{g})) = 188.83\)

The standard entropy change for the reaction is calculated using the formula:

\[ \Delta S^\circ = \sum S^\circ_{\text{products}} - \sum S^\circ_{\text{reactants}} \]

Substitute the values:

\[ \Delta S^\circ = \left[ S^\circ(\mathrm{MgCl}_{2}(\mathrm{s})) + 2 \times S^\circ(\mathrm{H}_{2}\mathrm{O}(\mathrm{g})) \right] - \left[ S^\circ(\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})) + 2 \times S^\circ(\mathrm{HCl}(\mathrm{g})) \right] \]

\[ \Delta S^\circ = \left[ 89.60 + 2 \times 188.83 \right] - \left[ 62.77 + 2 \times 186.90 \right] \]

\[ \Delta S^\circ = \left[ 89.60 + 377.66 \right] - \left[ 62.77 + 373.80 \right] \]

\[ \Delta S^\circ = 467.26 - 436.57 \]

\[ \Delta S^\circ = 30.69 \, \text{J/K} \]

Final Answer