Question

Consider the reaction 4PH3(g) → P4(g) + 6H2(g) At a particular point during the reaction, molecular hydrogen is being formed at the rate of 0.137 M/s.__(a) At what rate is P4 being produced? M/s (b) At what rate is PH3 being consumed? M/s

Answer 1

Answer:

The rate at which $P_4$ is being produced is 0.0228 M/s.

The rate at which $PH_3$ is being consumed is 0.0912 M/s.

Explanation:

$4PH_3\rightarrow P_4(g)+6H_2(g)$

Rate of the reaction : R

$R=\frac{-1}{4}\frac{d[PH_3]}{dt}=\frac{1}{6}\frac{d[H_2]}{dt}=\frac{1}{1}\frac{d[P_4]}{dt}$

The rate at which hydrogen is being formed = $\frac{d[H_2]}{dt}=0.137 M/s$

$R=\frac{1}{6}\frac{d[H_2]}{dt}$

$R=\frac{1}{6}\times 0.137 M/s=0.0228 M/s$

The rate at which $P_4$ is being produced:

$R=\frac{1}{1}\frac{d[P_4]}{dt}$

$0.0228 M/s=\frac{1}{1}\frac{d[P_4]}{dt}$

The rate at which $PH_3$ is being consumed :

$R=\frac{-1}{4}\frac{d[PH_3]}{dt}$

$0.0228 M/s\times 4=\frac{-1}{1}\frac{d[PH_3]}{dt}$

$\frac{-1}{1}\frac{d[PH_3]}{dt}=0.912 M/s$