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lara31 [8.8K]
2 years ago
5

Why does the dehydration of an alcohol more often use concentrated sulfuric acid, H 2 S O 4 HX2SOX4, as the acid catalyst rather

than dilute hydrochloric acid, H C l HCl
Chemistry
1 answer:
Dimas [21]2 years ago
4 0

KAnswer:

See explanation

Explanation:

It is more common to use H2SO4 for dehydration reaction rather than HCl because HCl contains a good nucleophile,the chloride ion.

Owing to the presence of the chloride ion, a substitution reaction involving the chloride ion may also proceed also thereby affecting the elimination reaction.

Also, concentrated H2SO4 is a very good drying agent thus, as long as it is used, the alcohol substrate is completely dehydrated to yield the alkene.

Note that HCl is not a dehydrating agent.

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At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---> +H2(g) + Br2 (g). If the initial partial pressures o
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Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

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The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

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