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erastovalidia [21]

3 years ago

14

Classify an element having the following ground state electron configuration as an alkali metal, alkaline earth metal, nonmetal,

halogen, transition metal, or noble gas.
a. [Ne]3s1
b. [Ne]3s23p3
c. [Ar]4s23d104p5
d. [Kr]5s24d1
e. [Kr]5s24d105p6

1 answer:

frutty [35]3 years ago

6 0

Explanation:

Alkali metal refers to group1 elements.

Alkali earth metal refers to group 2 elements.

Non metals refers to elements in grouos 4 to group 7.

Halogen refers to group 17 elements

Transition Metal refers to group 3 to group 12 elements

Noble gases refer to elements in group 18.

To obtain the group number from the electronic configuration, we calculate the total number of electrons in the principal quantum number (coefficient of the letters).

a. [Ne]3s1

Principal quantum number = 3

Number of electrons present = 1

This element belongs to group 1. It is an Alkali Metal.

b. [Ne]3s23p3

Principal quantum number = 3

Number of electrons present = 2 + 3 = 5

This element belongs to group 15 (5A). It is a Non metal

c. [Ar]4s23d104p5

Principal quantum number = 4

Number of electrons present = 2 + 5 = 7

This element belongs to group 17 (7A). It is an Halogen.

d. [Kr]5s24d1

This configuration belongs to the element yttrium and has an incomplete d shell. Hence it is a transition metal.

e. [Kr]5s24d105p6

Principal quantum number = 5

Number of electrons present = 2 + 6 = 8

This element belongs to group 18 (8A). It is a Noble gas.

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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 $(\Delta H^o)$ .

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

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

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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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