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creativ13 [48]
3 years ago
12

The following questions refer to the gas-phase decomposition of ethylene chloride. C2H5Cl ® products Experiment shows that the d

ecomposition is first order. The following data show kinetics information for this reaction: Time (s) ln [C2H5Cl] (M) 1.0 –1.625 2.0 –1.735 What is the half-life time for this reaction?
Chemistry
1 answer:
Inessa [10]3 years ago
3 0

Answer:

Half life = 6.3 seconds

Explanation:

C2H5Cl --> products

Experiment shows that the decomposition is first order.

Time (s) ln [C2H5Cl] (M)

1.0          –1.625

2.0         –1.735

The half-life of a first-order reaction is a constant that is related to the rate constant for the reaction:

t1/2 = 0.693/k.

To obtain the rate constant, k, we use the integral rate equation;

ln[A] = ln[A]o - kt

kt = ln[A]o - ln[A]

k = (ln[A]o - ln[A]) / t

k =  [–1.625  - (–1.735)] / 1

k = –1.625 + 1.735

k = 0.11

Half life, t1/2 = 0.693/ 0.11

Half life = 6.3 seconds

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One object could have more particles and greater total kinetic energy

Explanation:

The higher the temperature the more the particles. So, when we have high temperature, there is more particles interacting.

Temperature can simply be defined or gotten by taking the average of of the kinetic energy of the particles in the object that is the keywords here are TAKING THE AVERAGE KINETIC ENERGY

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So, when two particles of the same temperature have the different thermal energy it means that One object could have more particles and thus having greater total kinetic energy.

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I think it's B

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hoa [83]

Answer:

See explanation below

Explanation:

The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.

Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.

For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)

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