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MrRa [10]
3 years ago
13

One difference between strong bases and weak bases is 1. strong bases can neutralize more acid than weak bases. 2. strong bases

release more hydroxide ions than weak bases. 3. only strong bases can be amphiprotic. 4. weak bases dissociate completely; strong bases do not. 5. strong bases dissociate completely; weak bases do not.
Chemistry
1 answer:
Tema [17]3 years ago
6 0
One difference between strong bases and weak bases is that a strong bases dissociate completely while weak bases do not. A strong base is a compound that completely ionizes into metal and hydroxide ions when in a solution, on the other hand weak bases only partially ionizes to metal and hydroxide ions in solution. An example of a weak base is ammonia and an example of a strong base is sodium hydroxide.
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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

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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

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\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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