Question

For a gaseous reaction $A\to B+2C$ at ${250}^{\circ }$C, following data were observed. The partial pressure of the gases at 100th min will be (in 100x mm)

Time (min)	0	20	40	60
Total pressure (mm Hg)	400	500	587.6	664

Answer 1

$A\to B+2C$
t = 0 $a00$
t = t $a-xx2x$
$P_{total}\alpha a-x+2x\alpha a+2x$
a. i. $a\alpha 400$
$400+2x\alpha 500$
$x\alpha 50$
$k=\frac{2.303}{20}log\frac{400}{400-50}=0.00666mi{n}^{-1}$
ii. $a\alpha 400$
$400+2x\alpha 664$
$2x\alpha 264$
$x\alpha 132$
$k=\frac{2.303}{60}log\frac{400}{400-132}=0.00667mi{n}^{-1}$
Since the value of k is constant, hence first order.
b. 0.00667 = $\frac{2.303}{100}log\frac{400}{400-x}$
log x = 0.2896
x = 1.949 mm
Partial pressure of gases at 100 min = a + 2x
= 400 + 2 $\times$ 1.949
= 403.89 mm