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melisa1 [442]
4 years ago
15

During a constant-pressure process, the system releases heat to the surroundings. does the enthalpy of the system increase or de

crease during the process?
Chemistry
1 answer:
natta225 [31]4 years ago
5 0
The answer to this question would be: decrease
If the process is releasing heat to the surrounding, that means it should be an exothermic process. In the exothermic process, the process enthalpy difference will be minus because some of the energy is released as heat. That means the enthalpy of the system will be decreased.
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How to balance an unbalanced equation <br><br> C6H6+O2 yields CO2+H2O
storchak [24]
C6H6+15O2=12CO2+6H2O


4 0
4 years ago
Nitric oxide (NO) from car exhaust is a primary air pollutant. Calculate the equilibrium constant for the reaction
viva [34]

This problem is asking for the equilibrium constant at two different temperatures by describing the chemical equilibrium between gaseous nitrogen, oxygen and nitrogen monoxide at 25 °C and 1496 °C as the room temperature and the typical temperature inside the cylinders of a car's engine respectively:

N₂(g) + O₂(g) ⇄ 2 NO(g)

Thus, the calculated equilibrium constants turned out to be 6.19x10⁻³¹ and 9.87x10⁻⁵ at the aforementioned temperatures, respectively, according to the following work:

There is a relationship between the Gibbs free energy, enthalpy and entropy of the reaction, which leads to the equilibrium constant as shown below:

\Delta _rG=\Delta _rH-T\Delta _rS\\\\\Delta _rG=-RT ln(K)

Which means we can calculate the enthalpy and entropy of reaction and subsequently the Gibbs free energy and equilibrium constant. In such a way, we calculate these two as follows, according to the enthalpies of formation and standard entropies of N₂(g), O₂(g) and NO(g) since these are assumed constant along the temperature range:

\Delta _rH=2*90.25 kJ/mol - (0 kJ/mol+0 kJ/mol)=180.5kJ/mol\\\\\Delta _rS=2*(0.211 kJ/mol*K)-(0.192kJ/mol*K+0.205kJ/mol*K)=0.025kJ/mol*K

Then, we calculate the Gibbs free energy of reaction at both 25 °C and 1496 °C:

\Delta _rG_{25\°C}=180.5-(25+298.15)*0.025=172.42kJ/mol\\\\\Delta _rG_{1496\°C}=180.5-(1496+298.15)*0.025=135.65kJ/mol

And finally, the equilibrium constants derived from the general Gibbs equation and Gibbs free energies of reaction:

K=exp(-\frac{\Delta _rG}{RT} )\\\\K_{25\°C}=exp[-\frac{172420 J/mol}{(8.3145\frac{J}{mol*K})(298.15K)} ]=6.19x10^{-31}\\\\K_{1496\°C}=exp[-\frac{135650J/mol}{(8.3145\frac{J}{mol*K})(1769K)} ]=9.87x10^{-5}

Learn more:

  • (Gibbs free energy) brainly.com/question/15213613
4 0
3 years ago
explain why temperature is not as hot during the summer when a city is on a body of water (for example San Diego vs. Imperial Va
natita [175]

Answer:

because the water brings a cool breeze when the wind blows

Explanation:

4 0
3 years ago
Which structural formula does not represent benzene? a hexagon with nothing drawn inside it a hexagon with a dotted circle drawn
boyakko [2]
Answer:
            <span>A hexagon with nothing drawn inside it does not represent Benzene.

Explanation:

Below attached picture contains structures mentioned in given option.

Structure of Option-A is shown in Red Color.
</span>
Structure of Option-B is shown in Blue Color.

Structure of Option-C is shown in Green Color.

Structure of Option-D is shown in Orange Color.

Result:
          Except Option-A, all structures represent Benzene. Option-A in fact shows the structure of Cyclohexane.

5 0
3 years ago
Read 2 more answers
Combustion reactions typically involve all of the following except
cestrela7 [59]

Answer:

<h2><em><u>b. The production of a Hydrocarbon</u></em></h2>

Explanation:

In Combustion reaction usually occurs when a hydrocarbon reacts with oxygen to produce carbon dioxide and water. 

5 0
3 years ago
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