Question

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 25.0 s at 600°C,
calculate (a) the first-order rate constant for the
reaction and (b) the time required for 85% of the
phosphine to decompose.

Answer 1

a)

$$t_{1/2} = {ln2 \over k}$$

$$k = {ln2 \over t_{1/2}} = 0.028$$

b)

$$ln(0.85) = -kt = -0.028t$$

$$t = -{ln(0.85) \over 0.028} = 5.8 sec$$