Question

aces to access important values iseeded for this Isopropyl alcohol can dissociate into acetone and hydrogen: (CH3),CHOH(g) CH3),CO(g) +H;( ) At 179 °C, the equilibrium constant in terms of pressures, K, for this dissociation reaction is 0.444 (a) If 18.36 g of isopropyl alcohol is placed in a 11.90-L vessel and heated to 179°C, what is the partial pressure of acetone when equilibrium is attained? (b) What fraction of isopropyl alcohol is dissociated at equilibrium? Submit Antwer 2 question attempts remaining

Answer 1

PART (a):

No.of moles of isopropyl alcohol = (18.36 g) / (60.11 g/mol) = 0.30544 mol

Hence, From ideal gas equation, Partial pressure is given by:

$\small P=\frac{nRT}{V}=\frac{(0.30544\;mol)\times (0.0821\;L.atm/mol.K)\times (179+273\;K)}{(11.90\;L)}$

$\small \boldsymbol{\therefore P=0.952\;atm}$

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PART (b):

Let X be the change in pressure as it reaches the equilibrium.

X.X .Kp = 0.952 - x) (0.952-x) = 0.444

$\small \therefore \frac{x^{2}}{(0.952-x)}=0.444$

Solving for X, X = 0.465 atm

Therefore, Fraction of ispropyl alcohol at equilibrium :

$\small Fraction\;dissociated=\frac{0.465}{0.952}\times 100\%$

$\small \boldsymbol{Fraction\;dissociated=48.8\%}$