When a certain quantity of oxygen is ozonised in a suitable apparatus the volume decreased by 4ml. On addition of turpentine the volume further decreased by 8ml. All volumes were measured at the same temperature and pressure. From these data if formula of ozone is Ox then find x.
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P V0 = n R T n0 = number of moles of Oxygen, V0 = initial volume.
P & T are constant. V0 = k n0 where k is a constant
x O2 -> y Ozone
Let x moles of Oxygen be converted into y moles of O3
Volume V1 = k (n0 - x) + y k = k [ n0 - x + y ]
Change in volume = - 4 ml = - 0.004 = V1 - V0 = k [n0 - x + y] - k n0
so k ( x - y) = 0.004 --- equation 1
When turpentine is added, Ozone is completed dissolved in turpentine. Only remaining Oxygen O2 is present as gas.
Volume V2 = k (n0 - x)
change in volume = V2 - V1 = - 0.008 = k(n0 - x) - k (n0 - x + y)
So k y = 0.008 equation 2
From equation 1 and 2 , k x = 0.012
so k y / k x = y / x = 0.008 / 0.12 = 2/3
so 3 O2 ===> 2 molecules => 2 O3
P & T are constant. V0 = k n0 where k is a constant
x O2 -> y Ozone
Let x moles of Oxygen be converted into y moles of O3
Volume V1 = k (n0 - x) + y k = k [ n0 - x + y ]
Change in volume = - 4 ml = - 0.004 = V1 - V0 = k [n0 - x + y] - k n0
so k ( x - y) = 0.004 --- equation 1
When turpentine is added, Ozone is completed dissolved in turpentine. Only remaining Oxygen O2 is present as gas.
Volume V2 = k (n0 - x)
change in volume = V2 - V1 = - 0.008 = k(n0 - x) - k (n0 - x + y)
So k y = 0.008 equation 2
From equation 1 and 2 , k x = 0.012
so k y / k x = y / x = 0.008 / 0.12 = 2/3
so 3 O2 ===> 2 molecules => 2 O3
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