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katovenus [111]
3 years ago
10

Express the rate of the reaction in terms of the change in concentration of each of the reactants and products.

Chemistry
1 answer:
weeeeeb [17]3 years ago
4 0

Answer:

c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

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

Thus, the rate is given as:

rate=-\frac{1}{2} \frac{\Delta [HBr]}{\Delta t}=\frac{\Delta [Br_2]}{\Delta t} =\frac{\Delta [H_2]}{\Delta t}

It is necessary to remember that each concentration to time interval is divided into the stoichiometric coefficient, that is why HBr has a 1/2. Moreover, the concentration HBr is negative since it is a reactant and it has a negative rate due to its consumption.

Therefore, the answer is:

c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt

Best regards.

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How many sigma bonds in a single bond
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Read 2 more answers
At a certain temperature the rate of this reaction is first order in HI with a rate constant of :0.0632s
Shalnov [3]

Answer : The time taken for the reaction is, 28 s.

Explanation :

Expression for rate law for first order kinetics is given by :

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant  = 0.0632

t = time taken for the process  = ?

[A_o] = initial amount or concentration of the reactant  = 1.28 M

[A] = amount or concentration left time 't' = 1.28\times \frac{17}{100}=0.2176M

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

0.0632=\frac{2.303}{t}\log\frac{1.28}{0.2176}

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8 0
3 years ago
A student runs an experiment in the lab and then uses the data to prepare an Arrhenius plot of the natural log of the rate const
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Answer:

21.86582KJ

Explanation:

The graphical form of the Arrhenius equation is shown on the image attached. Remember that in the Arrhenius equation, we plot the rate constant against the inverse of temperature. The slope of this graph is the activation energy and its y intercept is the frequency factor.

Applying the equation if a straight line, y=mx +c, and comparing the given equation with the graphical form of the Arrhenius equation shown in the image attached, we obtain the activation energy of the reaction as shown.

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Explanation:

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