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dezoksy [38]
1 year ago
7

Use average bond energies to calculate ΔHrxn for the following hydrogenation reaction: H2C=CH2(g)+H2(g)→H3C−CH3(g)

Chemistry
1 answer:
marissa [1.9K]1 year ago
4 0

Answer:

The\Delta H_{rxn} of the given reaction is -129.6 kJ

Explanation:

The given chemical reaction is as follows.

H_{2}C=CH_{2}(g)+H_{2}(g)\rightarrow H_{3}C-CH_{3}(g)

Enthalpy of each reactant and products are as follows.

\Delta H_{C=C}\,=615.0\,kJ\,mol^{-1}

\Delta H _{H-H}\,=435.1\,kJ\,mol^{-1}

\Delta H _{C-C}\,=347.3\,kJ\,mol^{-1}

\Delta H _{C-H}\,=416.2\,kJ\,mol^{-1}

In the given chemical reaction involved two C-H bonds in the reactant side and one C-C bond in the product side therefore, the enthalpy of formation will be the negative.

\Delta H_{rxn}=-\Delta H_{C-C}-2\Delta H_{C-H}+\Delta H_{C=C}+\Delta H_{H-H}

=-347.4-2\times416.2+615.0+435.1

=-129.6 \,kJ

Therefore, The\Delta H_{rxn} of the given reaction is -129.6 kJ

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- Equilibrium shifts for slightly soluble compounds The reaction for the formation of a saturated solution of silver bromide (Ag
RideAnS [48]

Answer:

1) Favor formation of reactant

2) Favor formation of product

3) Favor formation of product

4) Favor formation of product

5) Favor formation of product

Explanation:

The reaction is AgBr(s)--->Ag^{+}(aq)+Br^{-}(aq)

The reaction is endothermic

1. Adding hydrobromic acid (HBr)

It will increase the amount of bromide ion. Bromide ion is one of the product so it will favor formation of reactant.

2. Lowering the concentration of bromide (Br-) ions

As we are decreasing the the concentration of product it will favor formation of more products

3. Heating the concentrations

If we are heating the mixture, it will increase the temperature and it will favor endothermic reaction. Thus will favor formation of product.

4. Adding solid silver bromide (AgBr)

If we are adding silver bromide it means we are increasing the concentration of reactant, it will favor formation of products

5. Removing silver (Ag+) ions

Removing silver ions means we are decreasing the concentration of product thus it will favor formation of products.

3 0
2 years ago
1.00 g of a compound is combusted in oxygen and found to give 3.14g of CO2 and 1.29 g of H2O. From these data we can tell thatA.
lora16 [44]

Answer:

the compound contains C, H, and some other element of unknownidentity, so we can’t calculate the empirical formula

Explanation:

Mass of CO2 obtained = 3.14 g

Hence number of moles of CO2 = 3.14g/44.0 g = 0.0714 mol

The mass of the carbon in the sample = 0.0714 mol × 12.0g/mol = 0.857 g

Mass of H2O obtained = 1.29 g

Hence number of moles of H2O = 1.29g/18.0 g = 0.0717 mol

The mass of the carbon in the sample = 0.0717 mol × 1g/mol = 0.0717 g

% by mass of carbon = 0.857/1 ×100 = 85.7 %

% by mass of hydrogen = 0.0717/1 × 100 = 7.17%

Mass of carbon and hydrogen = 85.7 + 7.17 = 92.87 %

Hence, there must be an unidentified element that accounts for (100 - 92.87) = 7.13% of the compound.

5 0
2 years ago
A researcher suspects that the pressure gauge on a 54.3-L gas cylinder containing nitric oxide is broken. An empty gas cylinder
mixer [17]

Answer:

Number of moles nitric acid in the cylinder is 400.539g/mol.

Explanation:

From the given,

Weight of empty gas cylinder W_{1}= 90.0 lb= 40823.3 gramsWeight of full cylinder[tex]W_{2} = 116.5 lb= 52843.511 gramsThe critical temperature = 287 KThe critical pressure 54.3 LMolar mass of nitric acid = [tex]M_{NO} = 30.01 g/mol

Number of moles nitric acid = n_{NO} =?

The mass of nitric acid in the cylinder = W_{NO}=W_{2}-W_{1}

=52843.511-40823.3 =12,020.2g

Number of moles of nitric acid =

\frac{Given\,mass}{Molar\,mass}=\frac{12,020.2}{30}=400.539g/mol

Therefore, number of moles nitric acid in the cylinder is 400.539g/mol.

3 0
2 years ago
After careful analysis, an electromagnetic wave is found to have a frequency of 7.8 x 10^6 Hz. What is the speed of the wave?
damaskus [11]

Answer:

 

Explanation:

4 0
2 years ago
Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the
Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by \eta.

The given data is as follows.

     \eta = 0.82,       T_{1} = (21 + 273) K = 294 K

     P_{1} = 200 kPa,     P_{2} = 1000 kPa

Therefore, calculate the final temperature as follows.

         \eta = \frac{T_{2} - T_{1}}{T_{2}}    

         0.82 = \frac{T_{2} - 294 K}{T_{2}}    

          T_{2} = 1633 K

Final temperature in degree celsius = (1633 - 273)^{o}C

                                                            = 1360^{o}C

Now, we will calculate the entropy as follows.

       \Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}

For 1 mole,  \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

It is known that for NH_{3} the value of C_{v} = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

     \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

                = 0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}

                = 0.0346 kJ/mol

or,             = 34.6 J/mol             (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

3 0
2 years ago
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