Decomposition of phosphine `(PH_(3))` at `120^(@)` C proceeds according to the equation `4PH_(3)(g) to P_(4)(g) + 6H_(2)(g)` It was found that the rea

Question

Decomposition of phosphine `(PH_(3))` at `120^(@)` C proceeds according to the equation `4PH_(3)(g) to P_(4)(g) + 6H_(2)(g)` It was found that the rea

Decomposition of phosphine `(PH_(3))` at `120^(@)` C proceeds according to the equation
`4PH_(3)(g) to P_(4)(g) + 6H_(2)(g)`
It was found that the reactioin follows the rate equation, Rate = `k[PH_(3)]`
The half life period of `PH_(3)` is 37.9 s at `120^(@)`C.
i) How much time will be required for `3//4` of `PH_(3)` to decompose? ltrbgt ii) What function of the original amount of `PH_(3)` will remain undecomposed after 1 minute?

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

i) For the first reaction,
`k=0.693/t_(1//2) = 0.693/(37.9S),t = 2.303/k log a/(a-x)` ltrbgt `t_(3//4) = (2.303 xx (37.9s))/(0.693) = 75.8s`
ii) `k=2.303/t log a/(a-x)`
`0.693/(37.9s) xx (60s)/(2.303) = 0.4764`
`a/(a-x) = "Antilog"0.4764 = 2.995`
If a=1, then (a-x) i.e. the fraction may be calculated as:
`1/(1-x) = 2.995 or (a-x) = 1/2.995 = 0.334`