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3 years ago

13

I have a chemistry Exam next class so I need help

1 answer:

zlopas [31]3 years ago

6 0

Answer:

1. Aluminium sulfide

$Al_{2}S_{3}$

2.Potassium oxide

$K_{2}O$

3.

$MgCl_{2}$ : Magnesium chloride

4.

HF : Hydrogen fluoride

5.

$H_{2}Se$ : Hydrogen selenide

6.

$CCl_{4}$ : Carbon tetrachloride

7.

$As_{2}O_{3}$ : Diarsenic Trioxide

8. Hydrobromic Acid : HBr

9.Chlorous Acid :

$HClO_{2}$

10.

$(NH_{4})Cr_{2}O_{7}$ : Ammonium dichromate

Explanation:

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Use the data given below to construct a Born-Haber cycle to determine the heat of formation of KCl. Δ H°(kJ) K(s) → K(g) 89 K(g)

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Explanation:

The net equation will be as follows.

$K(s) + Cl_{2}(g) \rightarrow KCl(s)$

So, we are required to find $\Delta H_{formation}$ for this reaction.

Therefore, steps involved for the above process are as follows.

Step 1: Convert K from solid state to gaseous state

$K(s) \rightarrow K(g)$ , $\Delta H_{1}$ = 89 kJ

Step 2: Ionization of gaseous K

$K(g) \rightarrow K^{+}(g) + e^{-}$ , $H_{2}$ = 418 KJ

Step 3: Dissociation of $Cl_{2}$ gas into chlorine atom .

$\frac{1}{2} Cl_{2}(g) \rightarrow Cl(g)$ , $\Delta H_{3} = \frac{244}{2}$ = 122 KJ

Step 4: Iozination of chlorine atom.

$Cl(g) + e^{-} \rightarro Cl^{-}(g)$ , $H_{4}$ = -349 KJ

Step 5: Add $K^{+}$ ion and $Cl^{-}$ ion formed above to get KCl .

$K^{+}(g) + Cl^{-}(g) \rightarrow KCl(s)$ , $H_{5}$ = -717 KJ

Now, using Born-Haber cycle, value of enthalpy of the formation is calculated as follows.

$\Delta H_{f} = \DeltaH_{1} + \Delta H_{2} + \Delta H_{3} + \Delta H_{4} + \Delta H_{5}$

= 89 + 418 + 122 - 349 - 717

= - 437 KJ/mol

Thus, we can conclude that the heat of formation of KCl is - 437 KJ/mol.

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