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gayaneshka [121]
4 years ago
9

Draw a Lewis structure for BCl3 and answer the following questions based on your drawing.

Chemistry
1 answer:
Georgia [21]4 years ago
4 0

Answer:

1. For the central boron atom:

A. The number of lone pairs = 0.

B. The number of single bonds = 3.

C. The number of double bonds = 0.

2. The central boron atom B. Has an incomplete octet.

Explanation:

Hello,

In this case, as shown on the attached picture containing the required Lewis structure of boron chloride, we can see that:

1. For the central boron atom:

A. The number of lone pairs = 0 since all the three valance electrons are bonded with the chlorine atoms.

B. The number of single bonds = 3 as each chlorine atom needs only one bond to obey the octet.

C. The number of double bonds = 0 considering the previous explanation.

2. The central boron atom B. Has an incomplete octet since it reaches six electrons only from its initial three and other three provided by the three chlorine atoms.

Regards.

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"Thermite" reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite react
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<u>Answer:</u> The given reaction is non-spontaneous in nature.

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]

For the given chemical reaction:

3Mg(s)+Cr_2O_3(s)\rightarrow 3MgO(s)+2Cr(s)

The equation for the entropy change of the above reaction is:

\Delta S^o_{rxn}=[(3\times \Delta S^o_{(MgO(s))})+(2\times \Delta S^o_{(Cr(s))})]-[(3\times \Delta S^o_{(Mg(s))})+(1\times \Delta S^o_{(Cr_2O_3(s))})]

We are given:

\Delta S^o_{(Mg(s))}=32.68J/K.mol\\\Delta S^o_{(Cr_2O_3(s))}=81.2J/K.mol\\\Delta S^o_{(MgO(s))}=26.94J/K.mol\\\Delta S^o_{(Cr(s))}=23.77J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(3\times (26.94))+(2\times (23.77))]-[(3\times (32.68))+(1\times (81.2))]\\\\\Delta S^o_{rxn}=-50.88J/K=-0.0509kJ/K.mol

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.

To calculate the standard Gibbs free energy of the reaction, we use the equation:

\Delta G^o=\Delta H^o-T\Delta S^o

where,

\Delta G^o = standard Gibbs free energy = ?

\Delta H^o = standard enthalpy change of the reaction = 665.1 kJ/mol

T = Temperature = 298.15 K

\Delta S^o = standard entropy change of the reaction = -0.0509 kJ/K.mol

Putting values in above equation, we get:

\Delta G^o=(665.1kJ/mol)-(298.15K\times (-0.0509kJ/K.mol))=680.27kJ/mol

As, the Gibbs free energy of the reaction is coming out to be positive, the reaction is non-spontaneous in nature.

Hence, the given reaction is non-spontaneous in nature.

4 0
4 years ago
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