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zysi [14]
3 years ago
15

Part b solid lithium metal and diatomic nitrogen gas react spontaneously to form a solid product. give the balanced chemical equ

ation (including phases) that describes this reaction. indicate the phases using abbreviation (s), (l), or (g) for solid, liquid, or gas, respectively.
Chemistry
1 answer:
dimaraw [331]3 years ago
3 0

Lithium metal which is an alkali metal belongs to the Group I of the periodic table reacts spontaneously with nitrogen, (which remains in gaseous form at room temperature) to form solid lithium nitride (Li_{3}N). The reaction can be shown as- 6Li (s) + N_{2}(g) = 2Li_{3}N(s). This is the only example of alkali nitride in chemistry. The other alkali nitrides are either not stable or does not exist in solid form at room temperature.

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Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang
Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

\Delta (H_{f})_{MnO_{2}} = -520.0 KJ/mole

\Delta (H_{f})_{Al_{2}O_{3}} = -1699.8 KJ/mole

The balanced chemical reaction is,

3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)

Formula used :

\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}

\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]

                        = (-3399.6) + (1560)

                        = -1839.6 KJ



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