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nataly862011 [7]
2 years ago
13

The vapor pressure of Substance X is measured at several temperatures:temperature (C) vapor pressure (atm)34 0.23644 0.29254 0.3

55Use this information to calculate the enthalpy of vaporization of X. Round your answer to 2 significant digits. Be sure your answer contains a correct unit symbol. Clears your work. Undoes your last action. Provides information about entering answers.
Chemistry
1 answer:
shusha [124]2 years ago
6 0

Answer:

ΔHv = 17.04 KJ/mol

Explanation:

T(°C)   T(K)      Pv(atm)         1/T(K)                LnPv

34      307      0.236       0.00325733      - 1.4439

44      317       0.292       0.0031545         - 1.2310

54      327      0.355       0.003058          - 1.0356

Clausius-Clapeyron:

  • δLnP/δT = ΔH/RT²

⇒ δLnP = ΔH/R (δT/T²)

∴ δT/T² = δ/δT(- 1/T )

⇒ δLnP/δT = - ΔH/R

Graphing: LnP vs 1/T

we get an ecuation that corresponds to a straight line:

y = - 2049.6x + 5.2331 ...... R² = 1

where the slope of this line is:

y = mx + b

⇒ m = - 2049.6 = - ΔH/R.....Clausius-Clapeyron

⇒ ΔH = (2049.6)(R)

∴ R = 8.314 E-3 KJ/mol.K

⇒ ΔHv = 17.04 KJ/mol

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Taya2010 [7]

Answer: The enthalpy of the reaction is -109 kJ

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

Fe_2O_3(s)+3CO(s)\rightarrow 2Fe(s)+3CO_2(g)   \Delta H=-28.0kJ  \times 3    

3Fe_2O_3(s)+9CO(s)\rightarrow 6Fe(s)+9CO_2(g)   \Delta H=-84.0kJ     (1)

3Fe(s)+4CO_2(s)\rightarrow 4CO(g)+Fe_3O_4(s) \Delta H=+12.5kJ  \times 2  

6Fe(s)+8CO_2(s)\rightarrow 8CO(g)+2Fe_3O_4(s) \Delta H=+25.0kJ     (2)

The final reaction is:

Subtracting (2) from (1):

3Fe_2O_3(s)+CO(g)\rightarrow CO_2(g)+2Fe_3O_4(s) \Delta H=-84.0-(+25.0)=-109kJ

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7 0
2 years ago
A 2 L bottle containing only air is sealed at a temperature of 22°C and a pressure of 0.982 atm. The bottle is placed in a freez
Degger [83]

Answer:

.997 atm

Explanation:

1. Find the combined gas law formula...

(P1V1/T1 = P2V2/T2)

2. Find our numbers...

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P2= ? (trying to find)

V1= 2 L

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- Note: always use Kelvin. To find Kelving add 273 to ___C.

3. Rearrange formula to fit problem...

(P2=P1V1T2/V2T1)

4. Fill in our values...

P2= .982 atm x 2 L x 270 K / 1.8 L x 295 K

5. Do the math and your answer should be...

.997 atm

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2 years ago
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<span>4p orbital</span>

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Explanation:

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