Isopropyl alcohol can dissociate into acetone and hydrogen according to the reac

Question

Isopropyl alcohol can dissociate into acetone and hydrogen according to the reaction below. At 179 degree C, the equilibrium constant for this dehydrogenation reaction is 0.444. If 0.166 moles of isopropyl alcohol is placed in a 10 L vessel and heated to 179 degree C, what is the partial pressure of acetone when equilibrium is attained? What fraction of isopropyl alcohol is dissociated? (Oxtoby 14.28) At 25 degree C, the equilibrium constant for the reaction below is 5.9 times 10^-13. Suppose a container is filled with nitrogen dioxide at an initial partial pressure of 0.89 atm. Calculate the partial pressures of all three gases after equilibrium is reached at this temperature. (Oxtoby 14.34) At 1330 K, germanium (II) oxide (Geo) and tungsten (VI) oxide (W_2 O_6) are both gases. The equilibrium constants for reactions i and are 7 times 10^3 and 3.8 times 10^4. Using this information, calculate K for reaction iii. (Oxtoby 14.20)

Explanation / Answer

Q1a.

K = 0.444

K = [P][H2]/[IP]

initially:

[IP] = 0.166/10 = 0.0166 M

[H2] = [P] = 0

in equilibrium

[IP] = 0.0166 -x

[H2] = [P] = 0+x

so:

0.444 = x*x/(0.0166 -x)

x = 0.016

so

[IP] = 0.0166 -x = 0.0166 -0.016 = 0.0006

[H2] = [P] = 0+x = 0.016

P-acetone = M*R*T = 0.016*0.082*(179+273) = 0.59302 atm

ii)

fraction dissociated = 0.016 /0.0166 = 0.96

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