Discussion Forum

The half-life of the first-order reaction $CH_{3}CHO(g)\rightarrow CH_{4}(g)+ CO(g)$ If the initial pressure of CH₃CHO (g) is 80 mm Hg and the total pressure at the end of 20 minutes is 120mmHg

Answer:

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= $\frac{ln2}{k}$ 

.'. t_1/2 = $\frac{ln2}{ln2}\times20$ = 20 min