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

Explanation:

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

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

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.

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$k=\frac{0.693}{35.0s}=0.0198s^{-1}$

The first order rate constant is $0.0198s^{-1}$

b) for completion of 95 % of reaction

$t=\frac{2.303}{k}\log\frac{100}{100-95}$

$t=\frac{2.303}{0.0198}\log\frac{100}{5}$

$t=151 s$

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