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Liula [17]
3 years ago
9

The reactivity of an alkali metal depends on the metals ability to

Chemistry
1 answer:
Klio2033 [76]3 years ago
4 0
<h3>Answer:</h3>

             The reactivity of an alkali metal depends on the metals ability to <u>loose an electron</u>.

<h3>Explanation:</h3>

                         Alkali metals are elements present in first group of the periodic table. Following are the metals from top to bottom;

                                               Lithium

                                               Sodium

                                               Potassium

                                               Rubidium

                                               Cesium

                                               Francium

These s-block elements are considered the most reactive metals in the periodic table because they have only one electron in their valence shell and readily loose that electron to form an ionic bond with non metal. The general electronic configuration of these metals is as follow,

                                                       nS¹

Also, the reactivity of these metals increases from top to bottom in the group as the ionization energy decreases due to increase in shielding effect respectively.

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Balance this nuclear reaction by supplying the missing nucleus: 249/98 Cf +__ --&gt; 263/106 Sg +4 1/0 n
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From these two reactions at 298 K, V2O3(s) + 3CO(g) → 2V(s) + 3CO2(g); ΔH° = 369.8 kJ; ΔS° = 8.3 J/K V2O5(s) + 2CO(g) → V2O3(s)
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Answer:

ΔG° = -133,1 kJ

Explanation:

For the reactions:

<em>(1) </em>V₂O₃(s) + 3CO(g) → 2V(s) + 3CO₂(g); ΔH° = 369,8 kJ; ΔS° = 8,3 J/K

<em>(2) </em>V₂O₅(s) + 2CO(g) → V₂O₃(s) + 2CO₂(g); ΔH° = –234,2 kJ; ΔS° = 0,2 J/K

By Hess's law it is possible to obtain the ΔH° and ΔS° of:

2V(s) + 5CO₂(g) → V₂O₅(s) + 5CO(g)

Substracting -(1)-(2), that means:

ΔH° = -369,8 kJ - (-234,2 kJ) = <em>-135,6 kJ</em>

ΔS° = - 8,3 J/K - 0,2 J/K =<em> -8,5 J/K</em>

Using: ΔG° = ΔH° - TΔS° at 298K

ΔG° = -135,6 kJ - 298K×-8,5x10⁻³kJ/K

<em>ΔG° = -133,1 kJ</em>

I hope it helps!

7 0
3 years ago
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