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Question

The decomposition of a compound P, at temperature T according to the equation
2P(g)4Q(g)+R(g)+S(l) is the first order reaction. After 30 minutes from the start of decomposition in a closed vessel, the total pressure developed is found to be 317 mm Hg and after a long period of time the total pressure observed to be 617 mm Hg. Calculate the total pressure of the vessel after 75 minute, if volume of liquid S is supposed to be negligible. Also calculate the time fraction t7/8?
Given : Vapour pressure of S(l) at temperature T = 32.5 mm Hg.

A
Pt=37.955mmHg,t7/8=39.996min
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B
Pt=379.55mmHg,t7/8=399.96min
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C
Pt=179.55mmHg,t7/8=199.96min
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D
None of these
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Solution

The correct option is B Pt=379.55mmHg,t7/8=399.96min
2P(g)4Q(g)+R(g)+S(l)
t=0 P0
t=30min. P0P2PP/2
t=- 2P0 P0/2
so P0P+2P+P/2=31732.5
i.e. P0+1.5P=284.5.........(i)
& 2.5P0=61732.5=584.5
soP0=233.8
P=33.8
k×30=ln233.8200k=0.0052
Att=75min
0.0052×75=ln233.8PP
PP=158.23P=75.57
PT=32.5+P0+1.5P=347.155+32.5
PT=379.65mmHg
(ii) 0.0052×t=ln8
t=399.89min.

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