A compound 'A' dissociate by two parallel first order paths atcertain temperature
`A(g) overset(k_(1)(min^(-1))rarr2B(g) k_(1)=6.93xx10^(-3) min^(-1)`
`A(g)overset(k_(2)(min^(-1))rarrC(g) k_(2)=6.93xx10^(-3) min^(-1)`
The reaction started whith 1 mole of pure A in 1 litre closed container with the initial pressure 2 atm what is the pressure (in atm ) developed in container after 50 minuts from start of experiment?

Options

A. `1.25`
B. `0.75`
C. `1.50`
D. `2.50`

Explanation

`{:(A,overset(k_(f))rarr,2B,,,A,overset(k_(2))rarr ,C),(a_(0)-x-y,,2x,,a_(0)-x-y,,y):}`
`(d[A])/(d t) = (k_(1) + k_(2))[A], ([B])/([C]) = (2k_(1))/(k_(2)) = (2x)/(y) = (k_(1))/(k_(2)) = (x)/(y) because k_(1) = k_(2) = x = y`
`(k_(1) + k_(2))t = ln""(A_(0))/(A_(t)), 2 xx 6.93 xx 10^(-3) xx 50 = ln""(A_(0))/(A_(t)) = [A]_(1) = ([A_(0)])/(2) = a_(0) x-y=a_(0)//2 (because x = y)`
`therefore x = a_(0)//4`. Total moles in container, `a_(0) -x - y + 2x + y = a_(0) + x = 1.25 a_(0)`
% increase in moles `rArr 25`, So final pressure `= 2 xx 1.25 = .5` atm