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Lina20 [59]
3 years ago
11

Consider the bonding properties of the three compounds

Chemistry
1 answer:
Vilka [71]3 years ago
3 0

Answer:

CO3^2- has the weakest C-O bond

Explanation:

Hybridization is a valence bond concept that has to do with the mixing of atomic orbitals to give hybrid atomic orbitals suitable for overlapping with orbitals of other atoms to form molecules. The hybridization of the carbon atom in CO, CO2 and CO3^2- are; sp, sp and sp2. This implies that there must be pi bonding in each of the species since sp and sp2 hybridized carbon atoms are known to lead to molecules possessing pi bonds.

In CO2, carbon in sp hybridized state is bonded to two oxygen atoms in sp2 hybridized state leading to a double bond between carbon and each oxygen atom. In CO, the carbon atom is sp hybridized while the oxygen atom is sp2 hybridized. CO3^2- contains an sp2 hybrized carbon atom in a trigonal planar geometry.

The relative bond lengths of the C-O bond in CO2, CO and CO3^2- is 116 pm, 112.8 pm and 136pm. Hence CO3^2- has the longest bond length. Remember that bond length is inversely proportional to the bond order. That is, the shorter the bond length, the greater the bond order. This implies that CO will have the greatest bond order because it shows the shortest bond length. It turns out that CO has a bond order of 3, CO2 has a bond order of 2 and CO3^2- has a bond order of 1.33. This is intermediate between the bond order of single and that of double C-O bond hence CO3^2- is represented using resonance structures.

The species with the longest C-O bond length also has the weakest C-O bond hence CO3^2- has the weakest C-O bond.

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Use the given data at 500 K to calculate ΔG°for the reaction
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Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

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Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

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

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

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