Question

10. 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.

a) Calculate the first-order rate constant for the reaction:

_______ s−1

b) Calculate the time required for 78.0 percent of the phosphine to decompose:

________ s

Answer 1

1)

Given:

Half life = 35 s

use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k

= 0.693/(half life)

= 0.693/(35)

= 1.98*10^-2 s-1

Answer: 1.98*10^-2 s-1

2)

we have:

[PH3]o = 100 [Let initial concentration be 100]

78.0 % has decomposed.

So, remaining is 22.0 %

[PH3] = 22.0

k = 1.98*10^-2 s-1

use integrated rate law for 1st order reaction

ln[PH3] = ln[PH3]o - k*t

ln(22) = ln(100) - 1.98*10^-2*t

3.091 = 4.605 - 1.98*10^-2*t

1.98*10^-2*t = 1.514

t = 76.47 s

Answer: 76.5 s