Consider the following gaseous reaction where initial pressure of A ( g ) is 200 torr A ( g ) arrow 3 B ( g )+ C ( g ) After time t, the pressure of system increased to 650 torr. Calculate the value of t (in seconds) (Given: rate constant of reaction, k =4.606× 10-3 sec -1, log 2=0.3 )

Question

Consider the following gaseous reaction where initial pressure of A ( g ) is 200 torr A ( g ) arrow 3 B ( g )+ C ( g ) After time t, the pressure of system increased to 650 torr. Calculate the value of t (in seconds) (Given: rate constant of reaction, k =4.606× 10-3 sec -1, log 2=0.3 ) (A) 600 (B) 300 (C) 50 (D) 5

Tardigrade Admin · Accepted Answer

A(g) longrightarrow 3 B(g)+C(g) at t=0 200 torr - - at t=t 200-p 3 p p 200-p+3 p+ p=650 p=150 torr t=(1/k) × 2.303 × log (200/200-p) =(1 × 2.303 × log 4/4.606 × 10-3)=300 sec

Q. Consider the following gaseous reaction where initial pressure of A(g) is 200 torr A(g)→3B(g)+C(g) After time t, the pressure of system increased to 650 torr. Calculate the value of t (in seconds) (Given: rate constant of reaction, k=4.606×10−3sec−1,log2=0.3 )

600

300

50

5

A(g)⟶3B(g)+C(g) at t=0200 torr −− at t=t200−p3pp 200−p+3p+p=650 p=150 torr t=k1​×2.303×log200−p200​ =4.606×10−31×2.303×log4​=300sec

$A (g) ⟶ 3 B (g) + C (g)$
at $t = 0200$ torr $- -$
at $t = t 200 - p 3 p p$
$200 - p + 3 p + p = 650$
$p = 150$ torr
$t = \frac{1}{k} \times 2.303 \times lo g \frac{200}{200 - p}$
$= \frac{1 \times 2.303 \times l o g 4}{4.606 \times 1 0 ^{- 3}} = 300 sec$