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Klio2033 [76]
3 years ago
10

Explain the trends you observed for bromine, chlorine, magnesium, sodium, and phosphorus. Explain whether they matched the predi

cted periodic trends.
Chemistry
2 answers:
notsponge [240]3 years ago
8 0
The atomic radii decreased across Period 3 and increased down Group 17. The ionic radii decreased across a period for the positive sodium and magnesium ions, increased for phosphorus, and then decreased again. The electron affinity and electronegativity increased across Period 3 and decreased down Group 17. The ionization energy increased across Period 3 and decreased down Group 17.
Explanation:

<span>--The atomic radii </span>diminished<span> across </span>amount three<span> and </span>raised<span> down </span>cluster seventeen.

<span>--The ionic radii </span>diminished<span> across an </span>amount<span> for the positive </span>atomic number 11<span> and </span>metal<span> ions, </span>raised<span> for phosphorus, </span>then diminished once more.

<span>--The </span>lepton<span> affinity and </span>negativity raised<span> across </span>amount three<span> and </span>diminished<span> down </span>cluster seventeen.

<span>--The ionization energy </span>raised<span> across </span>amount three<span> and </span>diminished<span> down </span>cluster seventeen<span>.</span>
Tanzania [10]3 years ago
6 0
The atomic radii decreased across Period 3 and increased down Group 17.The ionic radii decreased across a period for the positive sodium and magnesium ions, increased for phosphorus, and then decreased again.The electron affinity and electronegativity increased across Period 3 and decreased down Group 17.<span>The ionization energy increased across Period 3 and decreased down Group 17.</span>
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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)

\Delta S\ for\ graphite = 5.73 J/(K mol)

(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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