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SIZIF [17.4K]
3 years ago
14

Lewis structures cannot

Chemistry
1 answer:
mariarad [96]3 years ago
6 0
Lewis structure cannot show the numbers of valences
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Be sure to answer all parts. Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presenc
svlad2 [7]

Answer:

a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ,  ΔSºrxn =  0.113 kJ/K

b) At 753.55 ºC or higher

c )ΔG =  1.8 x 10⁴ J

    K = 8.2 x 10⁻²

Explanation:

a)                                 C6H5−CH2CH3  ⇒  C6H5−CH=CH2  + H₂

ΔHf kJ/mol                    -12.5                           103.8                      0

ΔGºf kJ/K                        119.7                         202.5                      0

Sº J/mol                          255                          238                      130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate  ΔHºrxn we have

ΔHºrxn  = ΔHfº C6H5−CH=CH2 +  ΔHfº H₂ - ΔHºfC6H5−CH2CH3

ΔHºrxn =     103.8 kJ + 0 kJ  - (-12.5 kJ)

ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 +  ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn=   202.5 kJ + 0 kJ - 119.7 kJ  = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn =  ΔHrxn -TΔS

we see that will happen when the term  TΔS  becomes greater than ΔHrxn since ΔS  is positive  , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that  ΔºHrxn and ΔSºrxn remain constant at the higher temperature  and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K  =  (1026.55 - 273 ) ºC = 753.55 ºC

c) Again we will use

                       ΔGrxn =  ΔHrxn -TΔS

to calculate ΔGrxn   with the assumption that ΔHº and ΔSºremain constant.

ΔG =  116.3 kJ - (600+273 K) x 0.113 kJ/K =  116.3 kJ - 873 K x 0.113 kJ/K

ΔG =  116.3 kJ - 98.6 kJ =  17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )

Now for solving for K, the equation to use is

ΔG = -RTlnK and solve for K

- ΔG / RT = lnK  ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K  x 873 K)) = 8.2 x 10⁻²

8 0
3 years ago
What is the difference between a purine and a pyrimidine?
GalinKa [24]

Answer: Adenine and guanine are the two purines and cytosine, thymine and uracil are the three pyrimidines. The main difference between purines and pyrimidines is that purines contain a sixmembered nitrogencontaining ring fused to an imidazole ring whereas pyrimidines contain only a sixmembered nitrogencontaining ring. They both are types or categories of nitrogen containing bases present in nuclei acids of DNA and RNA.

Purines are 2 Ring or Carbon Ring, Nitrogen containing bases. That consist of these 2 rings next placed next to each other. These examples include - Adenine and Guanine.

Pyrimidines are 1 or single Ring Nitrogen containing structures. There are 3 nitrogenous bases that are categorized as pyrimidines. Cytosine, Thymine, and Uracil.

4 0
2 years ago
20 Points!!<br><br> Need it as soon as possible.
Solnce55 [7]

Answer:

D.

Explanation:

The reaction is losing potential energy, which means that the reaction is losing that energy as heat. Exothermic is the loss of energy. Therefore it will be D.

8 0
3 years ago
Read 2 more answers
The half-life for the radioactive decay of ce−141 is 32.5 days. if a sample has an activity of 3.8 μci after 162.5 d have elapse
Umnica [9.8K]
Answer : 121.5 <span>μCi

Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.

We can use this formula;

</span>\frac{N}{ N_{0} } =  e^{-( \frac{0.693 X  T_{2} }{T_{1}})

3.8 / N_{0} = e^ ((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5  </span>μci
5 0
3 years ago
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Which adaptation is likely to increase the chances of survival of an animal in a rainforest?
LiRa [457]
I know I think it’s a A

6 0
3 years ago
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