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Llana [10]
3 years ago
13

Part B Write the balanced chemical equation for the neutralization reaction that occurs when an aqueous solution of hydrobromic

acid (HBr) is mixed with an aqueous solution of lithium hydroxide (LiOH). Indicate the physical states of the reactants and products using the abbreviations (s), (l), (g), or (aq) for solids, liquids, gases, or aqueous solutions, respectively. Express your answer as a chemical equation.
Chemistry
1 answer:
Doss [256]3 years ago
4 0

Answer:

HBr(aq) + LiOH(aq) → LiBr(aq) + H2O(l)

Explanation:

A neutralization reaction is a process in which an acid, aqeous HBr reacts completely with an appropriate amount of base, aqueous LiOH to produce salt, aqueous LiBr and water, liquid H2O only.

HBr(aq) + LiOH(aq) → LiBr(aq) + H2O(l)

Acid + base → Salt + Water.

During this reaction, the hydrogen ion, H+, from the HBr is neutralized by the hydroxide ion, OH-, from the LiOH to form the water molecule, H2O.

Thus, it is called a neutralization reaction.

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Consider the reaction. 2 HBr(g) ¡ H2(g) + Br2(g) a. Express the rate of the reaction in terms of the change in concentration of
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Answer :

(A) The rate expression will be:

Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}

(B) The average rate of the reaction during this time interval is, 0.00176 M/s

(C) The amount of Br₂ (in moles) formed is, 0.0396 mol

Explanation :

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The given rate of reaction is,

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

The expression for rate of reaction :

\text{Rate of disappearance of }HBr=-\frac{1}{2}\frac{d[HBr]}{dt}

\text{Rate of disappearance of }H_2=+\frac{d[H_2]}{dt}

\text{Rate of formation of }Br_2=+\frac{d[Br_2]}{dt}

<u>Part A:</u>

The rate expression will be:

Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}

<u>Part B:</u>

\text{Average rate}=-\frac{1}{2}\frac{d[HBr]}{dt}

\text{Average rate}=-\frac{1}{2}\frac{(0.512-0.600)M}{(25.0-0.0)s}

\text{Average rate}=0.00176M/s

The average rate of the reaction during this time interval is, 0.00176 M/s

<u>Part C:</u>

As we are given that the volume of the reaction vessel is 1.50 L.

\frac{d[Br_2]}{dt}=0.00176M/s

\frac{d[Br_2]}{15.0s}=0.00176M/s

[Br_2]=0.00176M/s\times 15.0s

[Br_2]=0.0264M

Now we have to determine the amount of Br₂ (in moles).

\text{Moles of }Br_2=\text{Concentration of }Br_2\times \text{Volume of solution}

\text{Moles of }Br_2=0.0264M\times 1.50L

\text{Moles of }Br_2=0.0396mol

The amount of Br₂ (in moles) formed is, 0.0396 mol

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