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

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.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

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