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Alika [10]
3 years ago
13

What is the value for ∆Soreaction for the following reaction, given the standard entropy values?

Chemistry
1 answer:
Vesnalui [34]3 years ago
6 0
The standard entropy for the substances are as follows:

C6H12O2(s) = -212 
<span>O2(g) = -205 </span>
<span>CO2(g) = -214 </span>
<span>H2O(l) = -70
</span>
We calculate the ∆S°r<span>eaction by the expression:

</span>∆S°rxn = ∆S°products - ∆S° reactants 
∆S°rxn = (212+6x205)-(6x214+6x70) 
∆S°rxn = -262 J/K ------> OPTION 3
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A 2.00-mol sample of hydrogen gas is heated at constant pressure from 294 K to 414 K. (a) Calculate the energy transferred to th
Furkat [3]

Answer:

a) The energy transferred is 6.91 kJ

b) The internal energy is 4.90 kJ

c) The work done on the gas is - 2.01 kJ

Explanation:

Step 1: Data given

Number of moles of hydrogen gas = 2.00 moles

Pressure = constant

Temperature is heated from 294 K to 414 K

Molar heat capacity of hydrogen gas = 28.8 J/mol*K

Step 2: Calculate the energy transferred to the gas by heat.

Q = n* Cp * ΔT

⇒with Q =the energy transferred

⇒with n = the number of moles = 2.00 moles

⇒with Cp = the Molar heat capacity of hydrogen gas = 28.8 J/mol*K

⇒ with ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K

Q = 2.00 * 28.8 * 120

Q = 6912 J = 6.91 kJ

Step 3: Calculate the increase in its internal energy.

ΔEint = n*Cv*ΔT

⇒with ΔEint = the increase in its internal energy.

⇒with n = the number of moles = 2.00 moles

⇒with Cv = The constant volume = 20.4 J/mol*K

⇒with  ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K

ΔEint = 2.00 * 20.4 * 120

ΔEint =4896 J = 4.90 kJ

Step 4: Calculate the work done on the gas.

Work done on the gas = -Q + ΔEint

W = -6.91 kJ + 4.90 kJ

W = -2.01 kJ

6 0
3 years ago
To test the effectiveness of a gunpowder mixture, 1 gram was exploded under controlled STP conditions, and the reaction chamber
Alekssandra [29.7K]

Answer:

1.3 × 10³ cm³

Explanation:

The gas occupies a volume of V₁ = 310 cm³ under standard temperature and pressure (STP), that is, T₁ = 273.15 K and P₁ = 1.0 atm. In order to find the volume V₂ under different conditions we can use the combined gas law formula.

\frac{P_{1}.V_{1}}{T_{1}} =\frac{P_{2}.V_{2}}{T_{2}} \\V_{2}=\frac{P_{1}.V_{1}.T_{2}}{T_{1}.P_{2}}=\frac{1.0atm\times 310cm^{3} \times 2473K }{273.15K \times 2.1atm} =1.3 \times 10^{3} cm^{3}

7 0
3 years ago
Which is the strongest acid listed in the table?
andrew11 [14]

Answer:

Hydrofluoric acid.

Explanation:

To know which of the acid is the strongest, let us determine the pka of each acid. This is illustrated below:

1. Acetic acid

Ka = 1.8x10^-5

pKa =..?

pKa = –logKa

pKa = –Log 1.8x10^-5

pKa = 4.74

2. Benzoic acid

Ka = 6.5x10^-5

pKa =..?

pKa = –logKa

pKa = –Log 6.5x10^-5

pKa = 4.18

3. Hydrofluoric acid.

Ka = 6.8x10^-4

pKa =..?

pKa = –logKa

pKa = –Log 6.8x10^-4

pKa = 3.17

4. Hypochlorous acid

Ka = 3.0x10^-8

pKa =..?

pKa = –logKa

pKa = –Log 3.0x10^-8

pKa = 7.52

Note: the smaller the pKa value, the stronger the acid.

The pka of the various acids as calculated above is given below:

Acid >>>>>>>>>>>>>>>>>> pKa

1. Acetic acid >>>>>>>>>> 4.74

2. Benzoic acid >>>>>>>> 4.18

3. Hydrofluoric acid >>>> 3.17

4. Hypochlorous acid >> 7.52

From the above illustration, we can see that hydrofluoric acid has the lowest pKa value. Therefore, hydrofluoric acid is the strongest among them.

6 0
3 years ago
To describe the length of a classroom, a student should use
Kamila [148]
The answer is D).....
8 0
3 years ago
Read 2 more answers
Compare melting point freezing point and boiling point
Tresset [83]
Melting point- the temperature at which a substance has changed from a solid to a liquid
freezing- the temperature at which a substance chanes from liquid to a solid
boiling point- the temperature at which a substance changes from a liquid to a gas phase
4 0
3 years ago
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