Chemistry, 19.03.2020 00:40 angel20202

Draw the product formed when cyclohexene is reacted with H2 in the presence of Pt. Note: If adding hydrogen atoms to a carbon atom, include all hydrogen atoms bonded to the carbon atom. Otherwise, the answer will be marked as incorrect.

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Answer from: dbhuggybearow6jng

It has been drawn and uploaded as an attachment. Please download it to see the structure.

Explanation:

The product formed as a result of the reaction of cyclohexene with H2 in presence of Pt (platinum) can be described as catalytic hydrogenation. Catalytic hydrogenation is defined as the process of hydrogen addition in the presence of a catalyst, which in this case is platinum.

Note that Cyclohexene (alkene) is a hydrocarbon molecule represented by the chemical formula, C6H10 .

It consists of a double bond. During the hydrogenation reaction, the alkene undergoes an addition reaction to give alkane which is a saturated hydrocarbon as the product.

The first step in order to derive the product is to draw the chemical structure of cyclohexene and identify the double bond present in it.

The final product can be derived by replacing the double bond with the single bond and satisfying all the valences of the carbon atom. The final product structure has been drawn and uploaded as an attachment. Please download it to see the structure.

Ans:

The structure of the cyclohexane thus, formed has been shown as follows with all the hydrogen atoms:

Draw the product formed when cyclohexene is reacted with H2 in the presence of Pt. Note: If adding h

Answer from: Quest

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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Draw the product formed when cyclohexene is reacted with H2 in the presence of Pt. Note: If adding h...

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