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ArbitrLikvidat [17]
3 years ago
10

please help with this ? Compare and contrast the compounds NH3 and PH3 in terms of bonding, molecular geometry (shape), and inte

rmolecular forces. Which substance would experience a stronger attraction between its molecules, and why?
Chemistry
2 answers:
nika2105 [10]3 years ago
6 0
 Both compounds are composed of covalent bonds and gaseous under room temperature and pressure. According to data, PH3 has a low boiling point than NH3, though the former being heavier than the later. This is due to intermolecular H- bonding in the later which is absent in the former. The oxidation state of phosphorous in PH3 is (- 3) as that for nitrogen in NH3. Both phosphine and ammonia react with oxygen to produce P2O5 gas and water, and N2 gas and water.
evablogger [386]3 years ago
3 0

Explanation: Both of these molecules are covalently bonded and are polar in nature. Since they are polar covalent molecules, they have dipole-dipole inter molecular forces of attraction.

In ammonia, N atom is the central atom and three H atoms are bonded to it and N atom has one lone pair of electrons due to which its molecular geometry is trigonal pyramidal.  Similarly, in phosphine, P is the central atom and three H atoms are bonded to it and it has one lone pair of electrons due to which its molecular geometry is trigonal pyramidal.

Nitrogen is more electron negative than phosphorous. Hydrogen boding is present in ammonia as hydrogen atoms are bonded to more electron negative nitrogen atom.

Due to the presence of hydrogen bonding, ammonia molecule experience stronger attraction forces as compared to phosphine.

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At 298 K, what is the Gibbs free energy change (ΔG) for the following reaction?
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Answer:

(a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

Explanation:

Given that,

Temperature = 298 K

Suppose, density of graphite is 2.25 g/cm³ and density of diamond is 3.51 g/cm³.

\Delta H\ for\ diamond = 1.897 kJ/mol

\Delta H\ for\ graphite = 0 kJ/mol

\Delta S\ for\ diamond = 2.38 J/(K mol)

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(a) We need to calculate the value of \Delta G for diamond

Using formula of Gibbs free energy change

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

Put the value into the formula

\Delta G= (1897-0)-298\times(2.38-5.73)

\Delta G=2895.3

\Delta G=2.895\ kJ

The Gibbs free energy  change is positive.

(b). When it is compressed isothermally from 1 atm to 1000 atm

We need to calculate the change of Gibbs free energy of diamond

Using formula of gibbs free energy

\Delta S=V\times\Delta P

\Delta S=\dfrac{m}{\rho}\times\Delta P

Put the value into the formula

\Delta S=\dfrac{12\times10^{-6}}{3.51}\times999\times10130

\Delta S=34.59\ J/mole

(c). Assuming that graphite and diamond are incompressible

We need to calculate the pressure

Using formula of Gibbs free energy

\beta= \Delta G_{g}+\Delta G+\Delta G_{d}

\beta=V(-\Delta P_{g})+\Delta G+V\Delta P_{d}

\beta=\Delta P(V_{d}-V_{g})+\Delta G

Put the value into the formula

0=\Delta P(\dfrac{12\times10^{-6}}{3.51}-\dfrac{12\times10^{-6}}{2.25})\times10130+2895.3

0=-0.0194\Delta P+2895.3

\Delta P=\dfrac{2895.3}{0.0194}

\Delta P=14924\ atm

(d). Here, C_{p}=0

So, The value of \Delta H and \Delta S at 900 k will be equal at 298 K

We need to calculate the Gibbs free energy of diamond relative to graphite

Using formula of Gibbs free energy

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

Put the value into the formula

\Delta G=(1897-0)-900\times(2.38-5.73)

\Delta G=4912\ J

Hence, (a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

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