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2 years ago

What do you think is the reason for an increasing atomic radius within one group?

Chemistry

2 answers:

Anna35 [415]2 years ago

7 0

Answer:

As you go down a group on the periodic table, you get more electron shells and therefore a larger atomic radius.

Marina86 [1]2 years ago

6 0

Periodic trends are patterns on the periodic table that
illustrate elements' characteristics.
Significant trends are atomic radius, metallic
reactivity, electronegativity, electron affinity, and
ionization energy.
Atomic radius and metallic reactivity trends:
They increase when they move down a group and left
across a period.
Electronegativity, electron affinity, and ionization
energy trends:
They increase when they move up a group and right
across a period.

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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 $(\Delta H^o)$ .

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

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

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 $(\Delta S^o)$ .

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

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