Answer:
Option C
Explanation:
$CH_{3}CHO(g)\rightarrow CH_{4}(g)+ CO(g)$
When t = 0 P0 0 0
When t = t P0 - P P P
.'. P0 - P + P + P = 120 mmHg
or, P0 + P = 120mmHg;
P = 120-80 = 40mmHg
k = $\frac{1}{t}ln\frac{P^{0}}{P^{0}-P}$ = $\frac{1}{20}ln\frac{80}{80-40}$ = $\frac{1}{20}ln2$
Again,t1/2= $\frac{ln2}{k}$
.'. t1/2 = $\frac{ln2}{ln2}\times20$ = 20 min