PCl 5(g) arrow PCl 3(g)+ Cl 2(g) In the above first order reaction the concentration of PCl 5 reduces from initial concentration 50 mol L-1 to 10 mol L-1 in 120 minutes at 300 ∼ K. The rate constant for the reaction at 300 K is x × 10-2 min -1. The value of x is [Given log 5=0.6989]

Question

PCl 5(g) arrow PCl 3(g)+ Cl 2(g) In the above first order reaction the concentration of PCl 5 reduces from initial concentration 50 mol L-1 to 10 mol L-1 in 120 minutes at 300 ∼ K. The rate constant for the reaction at 300 K is x × 10-2 min -1. The value of x is [Given log 5=0.6989] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/dcdbac713a321c60b05430bb1f067e5e-.png" /> ⇒ K=(2.303/t) log ([A0]/[At]) ⇒ K=(2.303/120) log (50/10) ⇒ K=(2.303/120) × 0.6989 =0.013413 min -1 =1.3413 × 10-2 min -1

Q. PCl5​(g)→PCl3​(g)+Cl2​(g) In the above first order reaction the concentration of PCl5​ reduces from initial concentration 50molL−1 to 10mol L−1 in 120 minutes at 300∼K. The rate constant for the reaction at 300K is x×10−2min−1. The value of x is _______ [Given log5=0.6989]

⇒K=t2.303​log[At​][A0​]​ ⇒K=1202.303​log1050​ ⇒K=1202.303​×0.6989 =0.013413min−1 =1.3413×10−2min−1

$\Rightarrow K = \frac{2.303}{t} lo g \frac{[ A _{0} ]}{[ A _{t} ]}$
$\Rightarrow K = \frac{2.303}{120} lo g \frac{50}{10}$
$\Rightarrow K = \frac{2.303}{120} \times 0.6989$
$= 0.013413 mi n^{- 1}$
$= 1.3413 \times 1 0^{- 2} mi n^{- 1}$