Question

Be sure to answer all parts. The thermal decomposition of phosphine (PH3) into phosphorus and molecular hydrogen is a first-order reaction: 4PH3(g) → P4(g) + 6H2(g) The half-life of the reaction is 35.0 s at 680°C.

Answer 1

The given question is incomplete. The complete question is :

The thermal decomposition of phosphine (PH3) into phosphorus and molecular hydrogen is a first-order reaction:
`4PH_3(g)\rightarrow P_4(g)+6H_2(g)` The half-life of the reaction is 35.0 s at 680°C. Calculate the first order rate constant.

Answer: a) The first order rate constant is
`0.0198s^(-1)`

b) The time after which 95% reactions gets completed is 151 seconds

Step-by-step explanation:

Expression for rate law for first order kinetics is given by:

`t=(2.303)/(k)\log(a)/(a-x)`

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

a) for finding the rate constant

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

`t_{(1)/(2)}=(0.693)/(k)`

`k=(0.693)/(35.0s)=0.0198s^(-1)`

The first order rate constant is
`0.0198s^(-1)`

b) for completion of 95 % of reaction

`t=(2.303)/(k)\log(100)/(100-95)`

`t=(2.303)/(0.0198)\log(100)/(5)`

`t=151 s`

The time after which 95% reactions gets completed is 151 seconds