Question

70) The thermal decomposition of phosphine (PHs) into phosphorus and molecular hydrogen is a first-order reaction: 4PH)(g)-→ P,(g) + 6H2(g) The half-life of the reaction is 30.0 s at 682°C. Calculate (a) the first-order rate constant for the reaction and (b) the time required for 85 percent of the phosphine to decompose.

can I please get help with this question. thanks

Answer 1

Here what I got,

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

Decomposition of Phosphine is a 1st Order reaction

HAlf-life of the reaction is equal to 30s at 682^o C

According to 1st order reaction

Half-Life of a First-order reaction (t_1/2 ) = $0.693/k1$

where k1 = 1st order rate-constant.

we also write $k1=0.693/t\texttt{1/2}$

(a) 1st Order rate constant = $0.693/30s$ =$0.0231s\texttt{-1}$

(b) let's assume initial concentration of Phosphine$(a)=100%$%

after decomposition of 85% of sample concentration $(a-x) =100-85=15%$ %

Time required to get decompose 85% of Phosphine = $1/k1\times\ln a/(a-x)$

=$1/0.0231\ln 100\div 15$$=82.123sec$