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Tresset [83]
2 years ago
7

Fe2O3(s) 2Al(s)Al2O3(s) 2Fe(s) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.73

moles of Fe2O3(s) react at standard conditions.
Chemistry
1 answer:
son4ous [18]2 years ago
6 0

Answer:

\Delta S=54.3\frac{J}{K}

Explanation:

Hello!

In this case, for the given reaction, we can write the equation to compute the entropy change as shown below:

\Delta s=2\Delta S_{Fe}+\Delta S_{Al_2O_3}-2\Delta S_{Al}-\Delta S_{Fe_2O_3}

Letting:

\Delta s_{Fe}=27.3\frac{J}{mol*K}\\\\ \Delta s_{Fe_2O_3}=84.4\frac{J}{mol*K}\\\\\Delta s_{Al}=28.3\frac{J}{mol*K}\\\\\Delta s_{Al_2O_3}=51.00\frac{J}{mol*K}

We obtain the entropy change per mole of Fe2O3(s):\Delta s=2*27.3\frac{J}{mol*K}+84.4\frac{J}{mol*K}-2*28.3\frac{J}{mol*K}-51.0\frac{J}{mol*K} \\\\\Delta s=31.4\frac{J}{mol*K}

Finally, the total entropy change when 1.73 moles of Fe2O3(s) react turns out:

\Delta s=31.4\frac{J}{mol*K}*1.73mol\\\\\Delta S=54.3\frac{J}{K}

Best regards!

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Explanation:

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After balancing the following reaction under acidic conditions, how many mole equivalents of water are required and on which sid
nordsb [41]

Answer:

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Explanation:

Hello,

In this case, we need to balance the given redox reaction in acidic media as shown below:

MnO_4^{1-} (aq) + Cl^{1-} (aq) \rightarrow  Mn^{2+} (aq) + Cl_2 (g)\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq) + Cl^{1-} (aq) \rightarrow  Mn^{2+} (aq) + Cl_2 (g)\\\\\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\\\\2Cl^{1-}\rightarrow Cl_2^0+2e^-\\\\2*[(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O]\\\\5*[2Cl^{1-}\rightarrow Cl_2^0+2e^-]\\\\\\\\2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\\\\10Cl^{1-}\rightarrow 5Cl_2^0+10e^-\\

Then, we add the half reactions:

2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10Cl^{1-} \rightarrow 2Mn^{2+}+8H_2O+5Cl_2^0

Thereby, we can see d. 8 moles of H2O on the product side.

Best regards.

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