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julia-pushkina [17]
3 years ago
5

Predict the products of the reaction below. that is, complete the right-hand side of the chemical equation. be sure your equatio

n is balanced and contains state symbols after every reactant and product. \chem;[h]\chem;[br](1[\chemwrong;[aq]]1)+[[\chem;[h]]_[2]]\chem;[o](1[\chemwrong;[l]]1)\chemarrow;[]
Chemistry
1 answer:
kkurt [141]3 years ago
8 0
The equation is as follow,

<span>                                  HBr </span>₍aq₎  +  H₂O ₍l₎    →

Solution:
             HBr being strong acid with Ka value of 1.0 × 10⁹. When HBr is added to water, water acts as a base and HBr acts as a acid. Water picks the proton (H⁺) from HBr and converts into Conjugate acid (H₃O⁺) ahile HBr is converted into Conjugate Base (Br⁻) after loosing proton. The equation for this reaction is as follow,

                      HBr ₍aq₎  +  H₂O ₍l₎    →    H₃O⁺ ₍aq₎  +  Br⁻ ₍aq₎
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Answer : The temperature for non-catalyzed reaction needed will be 456 K

Explanation :

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According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

Since, the rate for both the reaction are equal.

K_1=K_2

A\times e^{\frac{-Ea_1}{RT_1}}=A\times e^{\frac{-Ea_2}{RT_2}}

\frac{Ea_1}{T_1}=\frac{Ea_2}{T_2} ..........(1)

where,

Ea_1 = activation energy for non-catalyzed reaction = 75 kJ/mol

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T_1 =  temperature for non-catalyzed reaction = ?

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Now put all the given values in the above formula 1, we get:

\frac{Ea_1}{T_1}=\frac{Ea_2}{T_2}

\frac{75kJ/mol}{T_1}=\frac{49kJ/mol}{298K}

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