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galben [10]
3 years ago
10

Steam reforming of methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the start

ing point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills 1.5L a flask with of methane gas and 2.6 atm of water vapor at 47C, and when the mixture has come to equilibrium measures the partial pressure of carbon monoxide gas to be 1.4 atm.
Calculate the pressure equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.
Chemistry
1 answer:
lorasvet [3.4K]3 years ago
6 0

Answer:

Answer: Kp = 4.5

Explanation:

The balanced equation for the production of synthetic gas is

    CH4 +  H20 ⇒ CO + 3H2

Let x be the change in the concentration of each species at equilibrium

 CH4  +   H20   ⇔ CO  +  3H2

Initial                  0.60       2.6        0          0

Change               -x           -x          +x       +3x

Equilibrium      0.60-x      2.6-x       x        3x

Given that the equilibrium partial pressure of H2= 1.4 atm

then, 3x= 1.40

x= 1.4/3 = 0.466667

The equilibrium concentrations are

{CH4} = 0.60- x = 0.60 - 0.466667 = 0.133333atm

{H2O} = 2.60- x = 2.60 - 0.466667 = 2.133333atm

{CO} = x = 0.466667atm

{H2} = 1.4atm (given)

Kp  = <u>{CO}{H2}³</u>

        {CH4}{H2O}

Kp = <u>(0.466667)(1.4)³</u>

         (0.133333)(2.133333)

=  4.501875

Kp = 4.5

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A mixture of carbon dioxide and helium gases is maintained in a 7.91 L flask at a pressure of 1.42 atm and a temperature of 33 °
Crank

Answer:

The gas mixture contains 1.038 grams of helium

Explanation:

<u>Step 1:</u> Data given

Volume of the flask = 7.91 L

Total Pressure = 1.42 atm

Temperature = 33 °C

Mass of CO2 = 8.25 grams

Molar mass of CO2 = 44.01 g/mol

Molar mass of He = 4 g/mol

<u>Step 2</u>: Calculate total number moles of gas

p*V = n*R*T

⇒ p = the pressure = 1.42 atm

⇒ V = the volume = 7.91 L

⇒ n= the number of moles = TO BE DETERMINED

⇒ R = the gas constant = 0.08206 L* atm/K*mol

⇒ T = the temperature = 33 °C = 306 Kelvin

n = (p*V)/(R*T)

n = (1.42*7.91)/(0.08206 * 306)

n = 0.447 moles

<u>Step 3</u>: Calculate moles of CO2

Moles CO2 = mass CO2 / Molar mass CO2

Moles CO2 = 8.25 grams / 44.01 g/mol

Moles CO2 = 0.1875 moles

<u>Step 4:</u> Calculate moles of Helium

Moles Helium = total moles of gas - moles of CO2

Moles Helium = 0.447 - 0.1875 = 0.2595 moles of helium

<u>Step 5: </u> Calculate mass of helium

Mass of helium = moles of helium * molar mass of helium

Mass of helium = 0.2595 moles * 4 g/mol

Mass of helium = 1.038 grams

The gas mixture contains 1.038 grams of helium

8 0
3 years ago
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......................

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A straight chained hydrocarbon with this formula C5H10, has a triple paired C-C bond, has a double paired C-C bond, is essential
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A straight chain hydrocarbon with the formula C5 H10: has a double C-C bond
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A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g )
Aleksandr [31]

Answer:

(a) 7.11x10⁻⁴ M/s

(b) 2.56 mol.L⁻¹.h⁻¹

Explanation:

(a) The reaction is:

O₃(g) + NO(g) → O₂(g) + NO₂(g)   (1)

The reaction rate of equation (1) is given by:

rate = k*[O_{3}][NO]     (2)

<u>We have:</u>

k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹

[O₃]₀ = 2.35x10⁻⁶ M

[NO]₀ = 7.74x10⁻⁵ M

Hence, to find the inital reacion rate we will use equation (2):

rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s  

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

\frac{\Delta[NO_{2}]}{\Delta t} = rate

\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!                                

5 0
3 years ago
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