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Margaret [11]
1 year ago
13

The methane used to obtain H₂ for NH₃ manufacture is impure and usually contains other hydrocarbons, such as propane,C₃H₈. C₃H₈(

g) + 3H₂O(g) ⇄ 3CO(g) + 7H₂(g) Kp = 8.175×10¹⁵ at 1200KCO(g) + H₂O(g) ⇄ CO₂(g) + H₂(g) Kp = 0.6944 at 1200k(d) What percentage of the C₃H₈ remains unreacted?
Chemistry
1 answer:
Minchanka [31]1 year ago
4 0

According to the ideal gas law, partial pressure is inversely proportional to volume. It is also directly proportional to moles and temperature. At equilibrium in the following reaction at room temperature, the partial pressures of the gases are found to be PN2 = 0.094 atm, PH2 = 0.039 atm, and PNH3 = 0.003 atm.

<h3>Equilibrium partial pressures</h3>

The initial partial pressures of CO and water are 4.0 bar and 4.0 bar respectively.

The equilibrium partial pressures (in the bar) of CO, H2​O, CO2​, and H2​ are 4−p,4−p, and respectively.

Let p bar be the equilibrium partial pressure of hydrogen.

The expression for the equilibrium constant is

Kp​=PCO​PH2​O​PCO2​​PH2​​​=(4−p)(4−p)p×p​=0.1

p=1.264−0.316p

p=0.96 bar.

To learn more about equilibrium constant visit the link

brainly.com/question/10038290

#SPJ4

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As a result of the transfer of an electron from a less electronegative atom to a more electronegative atom,
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g Tibet (altitude above sea level is 29,028 ft) has an atmospheric pressure of 240. mm Hg. Calculate the boiling point of water
Marina CMI [18]

<u>Answer:</u> The boiling point of water in Tibet is 69.9°C

<u>Explanation:</u>

To calculate the boiling point of water in Tibet, we use the Clausius-Clayperon equation, which is:

\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

P_1 = initial pressure which is the pressure at normal boiling point = 1 atm = 760 mmHg      (Conversion factor:  1 atm = 760 mmHg)

P_2 = final pressure = 240. mmHg

\Delta H_{vap} = Heat of vaporization = 40.7 kJ/mol = 40700 J/mol     (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature or normal boiling point of water = 100^oC=[100+273]K=373K

T_2 = final temperature = ?

Putting values in above equation, we get:

\ln(\frac{240}{760})=\frac{40700J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{T_2}]\\\\-1.153=4895.36[\frac{T_2-373}{373T_2}]\\\\T_2=342.9K

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

T(K)=T(^oC)+273

342.9=T(^oC)+273\\T(^oC)=(342.9-273)=69.9^oC

Hence, the boiling point of water in Tibet is 69.9°C

3 0
3 years ago
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