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irinina [24]
3 years ago
10

General equilibrium problems. ICE type problems.a. Isopropyl alcohol can dissociate into acetone and hydrogen according to the r

eaction below.At 179 °C, the equilibrium constant for this dehydrogenation reaction is 0.444. i) If 0.166moles of isopropyl alcohol is placed in a 10 L vessel and heated to 179 °C, what is the partialpressure of acetone when equilibrium is attained
Chemistry
1 answer:
Pepsi [2]3 years ago
5 0

Answer:

Explanation:

In a gaseous reaction mixture partial pressure is proportion to mole of the gas concerned .

Pressure of the reactant gas from gas equation

PV = nRT

P = nRT / V

= .166 x .082 x ( 273+179) / 10

= .615 atm

C₃H₇OH   =     (CH₃)₂CO     +    H₂

before reaction moles in terms of pressure

.615                       0                      0

After reaction

.615 - x                     x                     x

.444 =  x² / ( .615 - x )

.273 - .444 x = x²

x² + .444 x - .273 = 0

x = .361 atm

So partial pressure of acetone is .361 atm at equilibrium.

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1) Write the chemical equation.

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Volume: 2.0 L.

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Pressure: 3.0 atm.

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Moles: <em>unknown</em>.

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<em>3.2- Plug in the known values and solve for n (moles).</em>

(3.0\text{ }atm)(2.0\text{ }L)=n*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)n=\frac{(3.0\text{ }atm)(2.0\text{ }L)}{(0.082057\text{ }L*atm*K^{-1}*mol^{-1})}=n=0.24\text{ }mol\text{ }CH_4

4) Moles of oxygen that reacted.

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mol\text{ }O_2=0.24\text{ }CH_4*\frac{2\text{ }mol\text{ }O_2}{1\text{ }mol\text{ }CH_4}=0.48\text{ }mol\text{ }O_2

5) Volume of oxygen required.

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Moles: 0.48 mol.

Temperature: 30 ºC = 303.15 K.

Pressure: 3.0 atm.

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Volume: <em>unknown</em>.

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PV=nRT

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